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#### Question 1:

(i) x increased by 12 is (x + 12).
(ii) y decreased by 7 is (- 7).
(iii) The difference of a and b, when a>b is (a - b).
(iv) The product of x and y is xy.
The sum of x and y is (x + y).
So, product of x and y added to their sum is xy + (x + y).
(v) One third of x is $\frac{x}{3}$.
The sum of a and b is (a + b).
$\therefore$ One-third of x multiplied by the sum of a and b =
(vi) 5 times x added to 7 times y =

(vii) Sum of x and the quotient of y by 5 is $\mathbf{x}\mathbf{+}\frac{\mathbf{y}}{\mathbf{5}}$.
(viii) x taken away from 4 is (4-x).
(ix) 2 less than the quotient of x by y is $\frac{\mathbf{x}}{\mathbf{y}}\mathbf{-}\mathbf{2}$.
(x) x multiplied by itself is $\mathrm{x}×\mathrm{x}={\mathbf{x}}^{\mathbf{2}}$.
(xi) Twice x increased by y is .
(xii)
Thrice x added to y squared is $\left(3×\mathrm{x}\right)+\left(\mathrm{y}×\mathrm{y}\right)=\mathbf{3}\mathbf{x}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}$.
(xiii) x minus twice y is $\mathrm{x}-\left(2×\mathrm{y}\right)=\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{y}$.
(xiv) x cubed less than y cubed is $\left(\mathrm{y}×\mathrm{y}×\mathrm{y}\right)-\left(\mathrm{x}×\mathrm{x}×\mathrm{x}\right)={\mathbf{y}}^{\mathbf{3}}\mathbf{-}{\mathbf{x}}^{\mathbf{3}}\mathbf{.}$
(xv) The quotient of x by 8 is multiplied by y is $\frac{\mathrm{x}}{8}×\mathrm{y}=\frac{\mathbf{xy}}{\mathbf{8}}$.

#### Question 2:

(i) x increased by 12 is (x + 12).
(ii) y decreased by 7 is (- 7).
(iii) The difference of a and b, when a>b is (a - b).
(iv) The product of x and y is xy.
The sum of x and y is (x + y).
So, product of x and y added to their sum is xy + (x + y).
(v) One third of x is $\frac{x}{3}$.
The sum of a and b is (a + b).
$\therefore$ One-third of x multiplied by the sum of a and b =
(vi) 5 times x added to 7 times y =

(vii) Sum of x and the quotient of y by 5 is $\mathbf{x}\mathbf{+}\frac{\mathbf{y}}{\mathbf{5}}$.
(viii) x taken away from 4 is (4-x).
(ix) 2 less than the quotient of x by y is $\frac{\mathbf{x}}{\mathbf{y}}\mathbf{-}\mathbf{2}$.
(x) x multiplied by itself is $\mathrm{x}×\mathrm{x}={\mathbf{x}}^{\mathbf{2}}$.
(xi) Twice x increased by y is .
(xii)
Thrice x added to y squared is $\left(3×\mathrm{x}\right)+\left(\mathrm{y}×\mathrm{y}\right)=\mathbf{3}\mathbf{x}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}$.
(xiii) x minus twice y is $\mathrm{x}-\left(2×\mathrm{y}\right)=\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{y}$.
(xiv) x cubed less than y cubed is $\left(\mathrm{y}×\mathrm{y}×\mathrm{y}\right)-\left(\mathrm{x}×\mathrm{x}×\mathrm{x}\right)={\mathbf{y}}^{\mathbf{3}}\mathbf{-}{\mathbf{x}}^{\mathbf{3}}\mathbf{.}$
(xv) The quotient of x by 8 is multiplied by y is $\frac{\mathrm{x}}{8}×\mathrm{y}=\frac{\mathbf{xy}}{\mathbf{8}}$.

Ranjit's score in English = 80 marks
Ranjit's score in Hindi = x marks
Total score in the two subjects = (Ranjit's score in English + Ranjit's score in Hindi)
∴ Total score in the two subjects = (80 + x) marks

#### Question 3:

Ranjit's score in English = 80 marks
Ranjit's score in Hindi = x marks
Total score in the two subjects = (Ranjit's score in English + Ranjit's score in Hindi)
∴ Total score in the two subjects = (80 + x) marks

(i) b × b × b × ... 15 times = ${\mathbf{b}}^{\mathbf{15}}$
(ii) y × y × y × ... 20 times = ${\mathbf{y}}^{\mathbf{20}}$
(iii) 14 × a × a × a × a × b × b × b =
(iv) 6 × x × x × y × y = $6×\left(\mathrm{x}×x\right)×\left(y×y\right)=\mathbf{6}{\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{2}}$
(v) 3 × z × z × z × y × y × x = $3×\left(\mathrm{z}×\mathrm{z}×z\right)×\left(\mathrm{y}×y\right)×x=\mathbf{3}{\mathbf{z}}^{\mathbf{3}}{\mathbf{y}}^{\mathbf{2}}\mathbf{x}$

#### Question 4:

(i) b × b × b × ... 15 times = ${\mathbf{b}}^{\mathbf{15}}$
(ii) y × y × y × ... 20 times = ${\mathbf{y}}^{\mathbf{20}}$
(iii) 14 × a × a × a × a × b × b × b =
(iv) 6 × x × x × y × y = $6×\left(\mathrm{x}×x\right)×\left(y×y\right)=\mathbf{6}{\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{2}}$
(v) 3 × z × z × z × y × y × x = $3×\left(\mathrm{z}×\mathrm{z}×z\right)×\left(\mathrm{y}×y\right)×x=\mathbf{3}{\mathbf{z}}^{\mathbf{3}}{\mathbf{y}}^{\mathbf{2}}\mathbf{x}$

(i) ${\mathrm{x}}^{2}{\mathrm{y}}^{4}=\left(\mathrm{x}×\mathrm{x}\right)×\left(\mathrm{y}×\mathrm{y}×\mathrm{y}×\mathrm{y}\right)=\mathbf{x}\mathbf{×}\mathbf{x}\mathbf{×}\mathbf{y}\mathbf{×}\mathbf{y}\mathbf{×}\mathbf{y}\mathbf{×}\mathbf{y}$
(ii) $6{\mathrm{y}}^{5}=6×\left(\mathrm{y}×\mathrm{y}×\mathrm{y}×\mathrm{y}×\mathrm{y}\right)=\mathbf{6}\mathbf{×}\mathbf{y}\mathbf{×}\mathbf{y}\mathbf{×}\mathbf{y}\mathbf{×}\mathbf{y}\mathbf{×}\mathbf{y}$
(iii) $9{\mathrm{xy}}^{2}\mathrm{z}=9×\mathrm{x}×\left(\mathrm{y}×\mathrm{y}\right)×\mathrm{z}=\mathbf{9}\mathbf{×}\mathbf{x}\mathbf{×}\mathbf{y}\mathbf{×}\mathbf{y}\mathbf{×}\mathbf{z}$
(iv) $10{\mathrm{a}}^{3}{\mathrm{b}}^{3}{\mathrm{c}}^{3}=10×\left(\mathrm{a}×\mathrm{a}×\mathrm{a}\right)×\left(\mathrm{b}×\mathrm{b}×\mathrm{b}\right)×\left(\mathrm{c}×\mathrm{c}×\mathrm{c}\right)=\mathbf{10}\mathbf{×}\mathbf{a}\mathbf{×}\mathbf{a}\mathbf{×}\mathbf{a}\mathbf{×}\mathbf{b}\mathbf{×}\mathbf{b}\mathbf{×}\mathbf{b}\mathbf{×}\mathbf{c}\mathbf{×}\mathbf{c}\mathbf{×}\mathbf{c}$

#### Question 1:

(i) ${\mathrm{x}}^{2}{\mathrm{y}}^{4}=\left(\mathrm{x}×\mathrm{x}\right)×\left(\mathrm{y}×\mathrm{y}×\mathrm{y}×\mathrm{y}\right)=\mathbf{x}\mathbf{×}\mathbf{x}\mathbf{×}\mathbf{y}\mathbf{×}\mathbf{y}\mathbf{×}\mathbf{y}\mathbf{×}\mathbf{y}$
(ii) $6{\mathrm{y}}^{5}=6×\left(\mathrm{y}×\mathrm{y}×\mathrm{y}×\mathrm{y}×\mathrm{y}\right)=\mathbf{6}\mathbf{×}\mathbf{y}\mathbf{×}\mathbf{y}\mathbf{×}\mathbf{y}\mathbf{×}\mathbf{y}\mathbf{×}\mathbf{y}$
(iii) $9{\mathrm{xy}}^{2}\mathrm{z}=9×\mathrm{x}×\left(\mathrm{y}×\mathrm{y}\right)×\mathrm{z}=\mathbf{9}\mathbf{×}\mathbf{x}\mathbf{×}\mathbf{y}\mathbf{×}\mathbf{y}\mathbf{×}\mathbf{z}$
(iv) $10{\mathrm{a}}^{3}{\mathrm{b}}^{3}{\mathrm{c}}^{3}=10×\left(\mathrm{a}×\mathrm{a}×\mathrm{a}\right)×\left(\mathrm{b}×\mathrm{b}×\mathrm{b}\right)×\left(\mathrm{c}×\mathrm{c}×\mathrm{c}\right)=\mathbf{10}\mathbf{×}\mathbf{a}\mathbf{×}\mathbf{a}\mathbf{×}\mathbf{a}\mathbf{×}\mathbf{b}\mathbf{×}\mathbf{b}\mathbf{×}\mathbf{b}\mathbf{×}\mathbf{c}\mathbf{×}\mathbf{c}\mathbf{×}\mathbf{c}$

(i) a+b
Substituting a = 2 and b = 3 in the given expression:
2+3 = 5

(ii) ${\mathrm{a}}^{2}+\mathrm{ab}$
Substituting a = 2 and b = 3 in the given expression:
$\left(2{\right)}^{2}+\left(2×3\right)=4+6\phantom{\rule{0ex}{0ex}}=10$

(iii) $\mathrm{ab}-{\mathrm{a}}^{2}$
Substituting a = 2 and b = 3 in the given expression:
$\left(2×3\right)-\left(2{\right)}^{2}=6-4\phantom{\rule{0ex}{0ex}}=2$

(iv) 2a-3b
Substituting a = 2 and b = 3 in the given expression:
$\left(2×2\right)-\left(3×3\right)=4-9\phantom{\rule{0ex}{0ex}}=-5$

(v) $5{\mathrm{a}}^{2}-2\mathrm{ab}$
Substituting a=2 and b=3 in the given expression:
$5×\left(2{\right)}^{2}-2×2×3=5×4-12=20-12\phantom{\rule{0ex}{0ex}}=8$

(vi) ${\mathrm{a}}^{3}-{\mathrm{b}}^{3}$
Substituting a=2 and b=3 in the given expression:
${2}^{3}-{3}^{3}=2×2×2-3×3×3=8-27\phantom{\rule{0ex}{0ex}}=-19$

#### Question 2:

(i) a+b
Substituting a = 2 and b = 3 in the given expression:
2+3 = 5

(ii) ${\mathrm{a}}^{2}+\mathrm{ab}$
Substituting a = 2 and b = 3 in the given expression:
$\left(2{\right)}^{2}+\left(2×3\right)=4+6\phantom{\rule{0ex}{0ex}}=10$

(iii) $\mathrm{ab}-{\mathrm{a}}^{2}$
Substituting a = 2 and b = 3 in the given expression:
$\left(2×3\right)-\left(2{\right)}^{2}=6-4\phantom{\rule{0ex}{0ex}}=2$

(iv) 2a-3b
Substituting a = 2 and b = 3 in the given expression:
$\left(2×2\right)-\left(3×3\right)=4-9\phantom{\rule{0ex}{0ex}}=-5$

(v) $5{\mathrm{a}}^{2}-2\mathrm{ab}$
Substituting a=2 and b=3 in the given expression:
$5×\left(2{\right)}^{2}-2×2×3=5×4-12=20-12\phantom{\rule{0ex}{0ex}}=8$

(vi) ${\mathrm{a}}^{3}-{\mathrm{b}}^{3}$
Substituting a=2 and b=3 in the given expression:
${2}^{3}-{3}^{3}=2×2×2-3×3×3=8-27\phantom{\rule{0ex}{0ex}}=-19$

(i) 3x-2y+4z
Substituting x = 1, y = 2 and z = 5 in the given expression:
$3×\left(1\right)-2×\left(2\right)+4×\left(5\right)=3-4+20\phantom{\rule{0ex}{0ex}}=19$

(ii)
Substituting x = 1, y = 2 and  z = 5 in the given expression:
${1}^{2}+{2}^{2}+{5}^{2}=\left(1×1\right)+\left(2×2\right)+\left(5×5\right)=1+4+25\phantom{\rule{0ex}{0ex}}=30$

(iii) $2{\mathrm{x}}^{2}-3{\mathrm{y}}^{2}+{\mathrm{z}}^{2}$
Substituting x = 1, y = 2 and z = 5 in the given expression:
$2×\left(1{\right)}^{2}-3×\left(2{\right)}^{2}+{5}^{2}=2×\left(1×1\right)-3×\left(2×2\right)+\left(5×5\right)=2-12+25\phantom{\rule{0ex}{0ex}}=15$

(iv) $\mathrm{xy}+\mathrm{yz}-\mathrm{zx}$
Substituting x = 1, y = 2 and z = 5 in the given expression:
$\left(1×2\right)+\left(2×5\right)-\left(5×1\right)=2+10-5\phantom{\rule{0ex}{0ex}}=7$

(v) $2{\mathrm{x}}^{2}\mathrm{y}-5\mathrm{yz}+{\mathrm{xy}}^{2}$
Substituting x = 1, y = 2 and z = 5 in the given expression:
$2×\left(1{\right)}^{2}×2-5×2×5+1×\left(2{\right)}^{2}=4-50+4\phantom{\rule{0ex}{0ex}}=-42$

(vi) ${x}^{3}-{y}^{3}-{z}^{3}\phantom{\rule{0ex}{0ex}}$
Substituting x = 1, y = 2 and z = 5 in the given expression:
${1}^{3}-{2}^{3}-{5}^{3}=\left(1×1×1\right)-\left(2×2×2\right)-\left(5×5×5\right)=1-8-125\phantom{\rule{0ex}{0ex}}=-132$

#### Question 3:

(i) 3x-2y+4z
Substituting x = 1, y = 2 and z = 5 in the given expression:
$3×\left(1\right)-2×\left(2\right)+4×\left(5\right)=3-4+20\phantom{\rule{0ex}{0ex}}=19$

(ii)
Substituting x = 1, y = 2 and  z = 5 in the given expression:
${1}^{2}+{2}^{2}+{5}^{2}=\left(1×1\right)+\left(2×2\right)+\left(5×5\right)=1+4+25\phantom{\rule{0ex}{0ex}}=30$

(iii) $2{\mathrm{x}}^{2}-3{\mathrm{y}}^{2}+{\mathrm{z}}^{2}$
Substituting x = 1, y = 2 and z = 5 in the given expression:
$2×\left(1{\right)}^{2}-3×\left(2{\right)}^{2}+{5}^{2}=2×\left(1×1\right)-3×\left(2×2\right)+\left(5×5\right)=2-12+25\phantom{\rule{0ex}{0ex}}=15$

(iv) $\mathrm{xy}+\mathrm{yz}-\mathrm{zx}$
Substituting x = 1, y = 2 and z = 5 in the given expression:
$\left(1×2\right)+\left(2×5\right)-\left(5×1\right)=2+10-5\phantom{\rule{0ex}{0ex}}=7$

(v) $2{\mathrm{x}}^{2}\mathrm{y}-5\mathrm{yz}+{\mathrm{xy}}^{2}$
Substituting x = 1, y = 2 and z = 5 in the given expression:
$2×\left(1{\right)}^{2}×2-5×2×5+1×\left(2{\right)}^{2}=4-50+4\phantom{\rule{0ex}{0ex}}=-42$

(vi) ${x}^{3}-{y}^{3}-{z}^{3}\phantom{\rule{0ex}{0ex}}$
Substituting x = 1, y = 2 and z = 5 in the given expression:
${1}^{3}-{2}^{3}-{5}^{3}=\left(1×1×1\right)-\left(2×2×2\right)-\left(5×5×5\right)=1-8-125\phantom{\rule{0ex}{0ex}}=-132$

(i) ${\mathrm{p}}^{2}+{\mathrm{q}}^{2}-{\mathrm{r}}^{2}$
Substituting p = -2, q = -1 and r = 3 in the given expression:
$\left(-2{\right)}^{2}+\left(-1{\right)}^{2}-\left(3{\right)}^{2}=\left(-2×-2\right)+\left(-1×-1\right)-\left(3×3\right)\phantom{\rule{0ex}{0ex}}⇒4+1-9=-4$

(ii) $2{\mathrm{p}}^{2}-{\mathrm{q}}^{2}+3{\mathrm{r}}^{2}$
Substituting p = -2, q = -1 and r = 3 in the given expression:

$2×\left(-2{\right)}^{2}-\left(-1{\right)}^{2}+3×\left(3{\right)}^{2}=2×\left(-2×-2\right)-\left(-1×-1\right)+3×\left(3×3\right)\phantom{\rule{0ex}{0ex}}⇒8-1+27=\mathbf{34}$

(iii) $\mathrm{p}-\mathrm{q}-\mathrm{r}$
Substituting p = -2, q = -1 and r = 3 in the given expression:
$\left(-2\right)-\left(-1\right)-\left(3\right)=-2+1-3\phantom{\rule{0ex}{0ex}}=-4$

(iv) ${\mathrm{p}}^{3}+{\mathrm{q}}^{3}+{\mathrm{r}}^{3}+3\mathrm{pqr}$
Substituting p = -2, q = -1 and r = 3 in the given expression:

$\left(-2{\right)}^{3}+\left(-1{\right)}^{3}+\left(3{\right)}^{3}+3×\left(-2×-1×3\right)\phantom{\rule{0ex}{0ex}}=\left(-2×-2×-2\right)+\left(-1×-1×-1\right)+\left(3×3×3\right)+3×\left(6\right)\phantom{\rule{0ex}{0ex}}=\left(-8\right)+\left(-1\right)+\left(27\right)+18\phantom{\rule{0ex}{0ex}}=36$

(v) $3{\mathrm{p}}^{2}\mathrm{q}+5{\mathrm{pq}}^{2}+2\mathrm{pqr}$
Substituting p = -2, q = -1 and r = 3 in the given expression:

$3×\left(-2{\right)}^{2}×\left(-1\right)+5×\left(-2\right)×\left(-1{\right)}^{2}+2×\left(-2×-1×3\right)\phantom{\rule{0ex}{0ex}}=3×\left(-2×-2\right)×\left(-1\right)+5×\left(-2\right)×\left(-1×-1\right)+2×\left(-2×-1×3\right)\phantom{\rule{0ex}{0ex}}=-12-10+12\phantom{\rule{0ex}{0ex}}=-10$

(vi) ${\mathrm{p}}^{4}+{\mathrm{q}}^{4}-{\mathrm{r}}^{4}$
Substituting p = -2, q = -1 and r = 3 in the given expression:
$\left(-2{\right)}^{4}+\left(-1{\right)}^{4}-\left(3{\right)}^{4}\phantom{\rule{0ex}{0ex}}=\left(-2×-2×-2×-2\right)+\left(-1×-1×-1×-1\right)-\left(3×3×3×3\right)\phantom{\rule{0ex}{0ex}}=16+1-81\phantom{\rule{0ex}{0ex}}=-64$

#### Question 4:

(i) ${\mathrm{p}}^{2}+{\mathrm{q}}^{2}-{\mathrm{r}}^{2}$
Substituting p = -2, q = -1 and r = 3 in the given expression:
$\left(-2{\right)}^{2}+\left(-1{\right)}^{2}-\left(3{\right)}^{2}=\left(-2×-2\right)+\left(-1×-1\right)-\left(3×3\right)\phantom{\rule{0ex}{0ex}}⇒4+1-9=-4$

(ii) $2{\mathrm{p}}^{2}-{\mathrm{q}}^{2}+3{\mathrm{r}}^{2}$
Substituting p = -2, q = -1 and r = 3 in the given expression:

$2×\left(-2{\right)}^{2}-\left(-1{\right)}^{2}+3×\left(3{\right)}^{2}=2×\left(-2×-2\right)-\left(-1×-1\right)+3×\left(3×3\right)\phantom{\rule{0ex}{0ex}}⇒8-1+27=\mathbf{34}$

(iii) $\mathrm{p}-\mathrm{q}-\mathrm{r}$
Substituting p = -2, q = -1 and r = 3 in the given expression:
$\left(-2\right)-\left(-1\right)-\left(3\right)=-2+1-3\phantom{\rule{0ex}{0ex}}=-4$

(iv) ${\mathrm{p}}^{3}+{\mathrm{q}}^{3}+{\mathrm{r}}^{3}+3\mathrm{pqr}$
Substituting p = -2, q = -1 and r = 3 in the given expression:

$\left(-2{\right)}^{3}+\left(-1{\right)}^{3}+\left(3{\right)}^{3}+3×\left(-2×-1×3\right)\phantom{\rule{0ex}{0ex}}=\left(-2×-2×-2\right)+\left(-1×-1×-1\right)+\left(3×3×3\right)+3×\left(6\right)\phantom{\rule{0ex}{0ex}}=\left(-8\right)+\left(-1\right)+\left(27\right)+18\phantom{\rule{0ex}{0ex}}=36$

(v) $3{\mathrm{p}}^{2}\mathrm{q}+5{\mathrm{pq}}^{2}+2\mathrm{pqr}$
Substituting p = -2, q = -1 and r = 3 in the given expression:

$3×\left(-2{\right)}^{2}×\left(-1\right)+5×\left(-2\right)×\left(-1{\right)}^{2}+2×\left(-2×-1×3\right)\phantom{\rule{0ex}{0ex}}=3×\left(-2×-2\right)×\left(-1\right)+5×\left(-2\right)×\left(-1×-1\right)+2×\left(-2×-1×3\right)\phantom{\rule{0ex}{0ex}}=-12-10+12\phantom{\rule{0ex}{0ex}}=-10$

(vi) ${\mathrm{p}}^{4}+{\mathrm{q}}^{4}-{\mathrm{r}}^{4}$
Substituting p = -2, q = -1 and r = 3 in the given expression:
$\left(-2{\right)}^{4}+\left(-1{\right)}^{4}-\left(3{\right)}^{4}\phantom{\rule{0ex}{0ex}}=\left(-2×-2×-2×-2\right)+\left(-1×-1×-1×-1\right)-\left(3×3×3×3\right)\phantom{\rule{0ex}{0ex}}=16+1-81\phantom{\rule{0ex}{0ex}}=-64$

(i) Coefficient of x in 13x is 13.
(ii) Coefficient of y in -5y is -5.
(iii) Coefficient of a in 6ab is 6b.
(iv) Coefficient of z in -7xz is -7x.
(v) Coefficient of p in -2pqr is -2qr.
(vi) Coefficient of y2 in 8xy2z is 8xz.
(vii) Coefficient of x3 in  x3 is 1.
(viii) Coefficient of x2 in -x2 is -1.

#### Question 5:

(i) Coefficient of x in 13x is 13.
(ii) Coefficient of y in -5y is -5.
(iii) Coefficient of a in 6ab is 6b.
(iv) Coefficient of z in -7xz is -7x.
(v) Coefficient of p in -2pqr is -2qr.
(vi) Coefficient of y2 in 8xy2z is 8xz.
(vii) Coefficient of x3 in  x3 is 1.
(viii) Coefficient of x2 in -x2 is -1.

(i) Numerical coefficient of ab is 1.
(ii) Numerical coefficient of -6bc is -6.
(iii) Numerical coefficient of 7xyz is 7.
(iv) Numerical coefficient of −2x3y2z is -2.

#### Question 6:

(i) Numerical coefficient of ab is 1.
(ii) Numerical coefficient of -6bc is -6.
(iii) Numerical coefficient of 7xyz is 7.
(iv) Numerical coefficient of −2x3y2z is -2.

A term of expression having no literal factors is called a constant term.
(i) In the expression 3x2 + 5x + 8, the constant term is 8.
(ii) In the expression 2x2 − 9, the constant term is -9.
(iii) In the expression  ${4y}^{2}-5y+\frac{3}{5}$, the constant term is $\frac{3}{5}$.
(iv) In the expression ${z}^{3}{-2z}^{2}+z-\frac{8}{3}$ , the constant term is $-\frac{8}{3}$.

#### Question 7:

A term of expression having no literal factors is called a constant term.
(i) In the expression 3x2 + 5x + 8, the constant term is 8.
(ii) In the expression 2x2 − 9, the constant term is -9.
(iii) In the expression  ${4y}^{2}-5y+\frac{3}{5}$, the constant term is $\frac{3}{5}$.
(iv) In the expression ${z}^{3}{-2z}^{2}+z-\frac{8}{3}$ , the constant term is $-\frac{8}{3}$.

The expressions given in (i), (iii), (vi) and (viii) contain only one term. So, each one of them is monomial.
The expressions given in (ii) and (ix) contain two terms. So, both of them are binomial.
The expressions given in (iv) and (v) contain  three terms. So, both of them are trinomial.
The expression given in (vii) contains four terms. So, it does not represents any of the given types.

#### Question 8:

The expressions given in (i), (iii), (vi) and (viii) contain only one term. So, each one of them is monomial.
The expressions given in (ii) and (ix) contain two terms. So, both of them are binomial.
The expressions given in (iv) and (v) contain  three terms. So, both of them are trinomial.
The expression given in (vii) contains four terms. So, it does not represents any of the given types.

(i) Expression 4x5 − 6y4 + 7x2y − 9 has four terms, namely 4x5 ,-6y4 , 7x2y and -9.
(ii) Expression 9x3 − 5z4 + 7z3 y − xyz has four terms, namely 9x3 , -5z4 , 7z3 y and -xyz.

#### Question 9:

(i) Expression 4x5 − 6y4 + 7x2y − 9 has four terms, namely 4x5 ,-6y4 , 7x2y and -9.
(ii) Expression 9x3 − 5z4 + 7z3 y − xyz has four terms, namely 9x3 , -5z4 , 7z3 y and -xyz.

The terms that have same literals are called like terms.
(i)  a2 and 2a2 are like terms.
(ii)  are like terms.
(iii) −2xy2 and 5y2x are like terms.
(iv) ab2c , acb2 , b2ac and cab2 are like terms.

#### Question 1:

The terms that have same literals are called like terms.
(i)  a2 and 2a2 are like terms.
(ii)  are like terms.
(iii) −2xy2 and 5y2x are like terms.
(iv) ab2c , acb2 , b2ac and cab2 are like terms.

(i) Required sum = 3x + 7x
= (3+7)x = 10x

(ii) Required sum = 7y +(−9y)
= (7-9)y = -2y

(iii) Required sum = 2xy +5xy + (−xy)
= (2+5-1)xy = 6xy

(iv) Required sum = 3x+2y

(v) Required sum = 2x2 + (− 3x2) + 7x2
=(2-3+7)x2 = 6x2

(vi)Required sum =  7xyz + (− 5xyz) + 9xyz + (−8xyz)
= (7-5+9-8)xyz = 3xyz

(vii) Required sum = 6a3 +(− 4a3) + 10a3 +( −8a3)
=(6-4+10-8)a3 = 4a3

(viii) Required sum = x2 − a2 + (−5x2 + 2a2) +( −4x2 + 4a2 )
Rearranging and collecting the like terms =  x2 -5x2 -4x2 -a2 + 2a2 +4a2
= (1-5-4)x2 +(-1+2+4)a2
= -8x2 + 5a2

#### Question 2:

(i) Required sum = 3x + 7x
= (3+7)x = 10x

(ii) Required sum = 7y +(−9y)
= (7-9)y = -2y

(iii) Required sum = 2xy +5xy + (−xy)
= (2+5-1)xy = 6xy

(iv) Required sum = 3x+2y

(v) Required sum = 2x2 + (− 3x2) + 7x2
=(2-3+7)x2 = 6x2

(vi)Required sum =  7xyz + (− 5xyz) + 9xyz + (−8xyz)
= (7-5+9-8)xyz = 3xyz

(vii) Required sum = 6a3 +(− 4a3) + 10a3 +( −8a3)
=(6-4+10-8)a3 = 4a3

(viii) Required sum = x2 − a2 + (−5x2 + 2a2) +( −4x2 + 4a2 )
Rearranging and collecting the like terms =  x2 -5x2 -4x2 -a2 + 2a2 +4a2
= (1-5-4)x2 +(-1+2+4)a2
= -8x2 + 5a2

(i)

(ii)

(iii)

(iv)

#### Question 3:

(i)

(ii)

(iii)

(iv)

(i) Sum of the given expressions
= (3a − 2b + 5c)+(2a + 5b − 7c)+ (− a − b + c)
Rearranging and collecting the like terms
= 3a+2a-a-2b+5b-b+5c-7c+c
= (3+2-1)a + (-2+5-1)b + (5-7+1)c
= 4a+2b-c

(ii) Sum of the given expressions
= (8a − 6ab + 5b) + (−6a − ab − 8b) + (−4a + 2ab + 3b)
Rearranging and collecting the like terms
=(8−6−4)a + (- 6 −1+2)ab + (5− 8+ 3)b
= -2a-5ab+0 = -2a - 5ab

(iii) Sum of the given expressions
= (2x3 − 3x2 + 7x − 8) + (−5x3 + 2x2 − 4x + 1) + ( 3 − 6x + 5x2 − x3 )
Rearranging and collecting the like terms
=2x3−5x3 − x3 − 3x2 + 2x2 + 5x2 +7x-4x-6x-8+1+3
= (2-5-1)x3 +(-3+2+5)x2+(7-4-6)x-4
= -4x3 +4x2-3x-4

(iv) Sum of the given expressions
= (2x2 − 8xy + 7y2 − 8xy2)+( 2xy2 + 6xy − y2 + 3x2)+( 4y2 − xy − x2 + xy2 )
Rearranging and collecting the like terms
= 2x2 +3x2 − x2  + 7y2 − y2 +4y2 − 8xy + 6xy − xy− 8xy2 +2xy2 + xy2
= (2 +3− 1)x2  + (7 − 1 +4)y2 + (-8 + 6 −1)xy + (− 8 +2 +1)xy2
= 4x2  + 10y2 − 3xy -5xy2

(v) Sum of the given expressions
= (x3 + y3 − z3 + 3xyz)+(− x3 + y3 + z3 − 6xyz)+(x3 − y3 − z3 − 8xyz)
Rearranging and collecting the like terms
= x3 -x3 + x3 + y3 + y3 − y3 -z3 + z3 − z3 + 3xyz-6xyz-8xyz
= (1-1+1)x3 + (1+1-1)y3 + (-1+1-1)z3 +(3-6-8)xyz
= x3 + y3 − z3 -11xyz

(vi) Sum of the given expressions
= (2 + xx2 + 6x3)+(−6 −2x + 4x2 −3x3)+( 2 + x2)+( 3 − x3 + 4x − 2x2 )
Rearranging and collecting the like terms
= 6x3 −3x3x3x2 +4x2+ x2− 2x2+ x −2x+ 4x+2-6+2+3
=  (6-3-1)x3+(-1+4+1-2)x2+(1-2+4)x+1
= 2x3+2x2+3x+1

#### Question 4:

(i) Sum of the given expressions
= (3a − 2b + 5c)+(2a + 5b − 7c)+ (− a − b + c)
Rearranging and collecting the like terms
= 3a+2a-a-2b+5b-b+5c-7c+c
= (3+2-1)a + (-2+5-1)b + (5-7+1)c
= 4a+2b-c

(ii) Sum of the given expressions
= (8a − 6ab + 5b) + (−6a − ab − 8b) + (−4a + 2ab + 3b)
Rearranging and collecting the like terms
=(8−6−4)a + (- 6 −1+2)ab + (5− 8+ 3)b
= -2a-5ab+0 = -2a - 5ab

(iii) Sum of the given expressions
= (2x3 − 3x2 + 7x − 8) + (−5x3 + 2x2 − 4x + 1) + ( 3 − 6x + 5x2 − x3 )
Rearranging and collecting the like terms
=2x3−5x3 − x3 − 3x2 + 2x2 + 5x2 +7x-4x-6x-8+1+3
= (2-5-1)x3 +(-3+2+5)x2+(7-4-6)x-4
= -4x3 +4x2-3x-4

(iv) Sum of the given expressions
= (2x2 − 8xy + 7y2 − 8xy2)+( 2xy2 + 6xy − y2 + 3x2)+( 4y2 − xy − x2 + xy2 )
Rearranging and collecting the like terms
= 2x2 +3x2 − x2  + 7y2 − y2 +4y2 − 8xy + 6xy − xy− 8xy2 +2xy2 + xy2
= (2 +3− 1)x2  + (7 − 1 +4)y2 + (-8 + 6 −1)xy + (− 8 +2 +1)xy2
= 4x2  + 10y2 − 3xy -5xy2

(v) Sum of the given expressions
= (x3 + y3 − z3 + 3xyz)+(− x3 + y3 + z3 − 6xyz)+(x3 − y3 − z3 − 8xyz)
Rearranging and collecting the like terms
= x3 -x3 + x3 + y3 + y3 − y3 -z3 + z3 − z3 + 3xyz-6xyz-8xyz
= (1-1+1)x3 + (1+1-1)y3 + (-1+1-1)z3 +(3-6-8)xyz
= x3 + y3 − z3 -11xyz

(vi) Sum of the given expressions
= (2 + xx2 + 6x3)+(−6 −2x + 4x2 −3x3)+( 2 + x2)+( 3 − x3 + 4x − 2x2 )
Rearranging and collecting the like terms
= 6x3 −3x3x3x2 +4x2+ x2− 2x2+ x −2x+ 4x+2-6+2+3
=  (6-3-1)x3+(-1+4+1-2)x2+(1-2+4)x+1
= 2x3+2x2+3x+1

Change the sign of each term of the expression that is to be subtracted and then add.

(i) Term to be subtracted = 5x
Changing the sign of each term of the expression gives -5x.

2x+(-5x) = 2x-5x
= (2-5)x
= -3x

(ii)  Term to be subtracted = -xy
Changing the sign of each term of the expression gives xy.

6xy+xy
= (6+1)xy
= 7xy

(iii) Term to be subtracted = 3a
Changing the sign of each term of the expression gives -3a.

5b+(-3a)
= 5b-3a

(iv) Term to be subtracted = -7x
Changing the sign of each term of the expression gives 7x.
9y+7x

(v) Term to be subtracted = 10x2
Changing the sign of each term of the expression gives -10x2.
−7x2 + (-10x2) = −7x2 −10x2
= (−7−10)x2
= −17x2

(vi) Term to be subtracted = a2 − b2
Changing the sign of each term of the expression gives -a2 + b2.
b2 − a2 + (-a2 + b2) = b2 − a2 -a2 + b2
= (1+1)b2  +(−1-1) a2
= 2b2 − 2a2

#### Question 5:

Change the sign of each term of the expression that is to be subtracted and then add.

(i) Term to be subtracted = 5x
Changing the sign of each term of the expression gives -5x.

2x+(-5x) = 2x-5x
= (2-5)x
= -3x

(ii)  Term to be subtracted = -xy
Changing the sign of each term of the expression gives xy.

6xy+xy
= (6+1)xy
= 7xy

(iii) Term to be subtracted = 3a
Changing the sign of each term of the expression gives -3a.

5b+(-3a)
= 5b-3a

(iv) Term to be subtracted = -7x
Changing the sign of each term of the expression gives 7x.
9y+7x

(v) Term to be subtracted = 10x2
Changing the sign of each term of the expression gives -10x2.
−7x2 + (-10x2) = −7x2 −10x2
= (−7−10)x2
= −17x2

(vi) Term to be subtracted = a2 − b2
Changing the sign of each term of the expression gives -a2 + b2.
b2 − a2 + (-a2 + b2) = b2 − a2 -a2 + b2
= (1+1)b2  +(−1-1) a2
= 2b2 − 2a2

Change the sign of each term of the expression that is to be subtracted and then add.

(i) Term to be subtracted = 5a + 7b − 2c
Changing the sign of each term of the expression gives -5a -7b + 2c.
(3a − 7b + 4c)+(-5a -7b + 2c ) = 3a − 7b + 4c-5a -7b + 2c
= (3-5)a+( − 7-7)b + (4+2)c
= -2a − 14b + 6c

(ii) Term to be subtracted = a − 2b − 3c
Changing the sign of each term of the expression gives -a +2b + 3c.
(−2a + 5b − 4c)+(-a +2b + 3c ) = −2a + 5b − 4c-a +2b + 3c
= (−2-1)a + (5+2)b +(−4+3)c
= −3a + 7b − c

(iii) Term to be subtracted = 5x2 − 3xy + y2
Changing the sign of each term of the expression gives  -5x2 + 3xy - y2.

(7x2 − 2xy − 4y2)+(-5x2 + 3xy - y2) = 7x2 − 2xy − 4y2-5x2 + 3xy - y2
= (7-5)x2 +(−2+3)xy +(−4-1)y2
= 2x2 +xy − 5y2

(iv) Term to be subtracted = 6x3 − 7x2 + 5x − 3
Changing the sign of each term of the expression gives  -6x3 + 7x2 - 5x + 3.
(4 − 5x + 6x2 − 8x3)+(-6x3 + 7x2 - 5x + 3) = 4 − 5x + 6x2 − 8x3-6x3 + 7x2 - 5x + 3
= (-8-6)x3 +(6+7)x2 +(-5- 5)x + 7
= -14x3 + 13x2 - 10x + 7

(v) Term to be subtracted = x3 + 2x2y + 6xy2 − y3
Changing the sign of each term of the expression gives  -x3 - 2x2y - 6xy2 + y3.
(y3 − 3xy2 − 4x2y)+(-x3 - 2x2y - 6xy2 + y3) = y3 − 3xy2 − 4x2y-x3 - 2x2y - 6xy2 + y3
= -x3 +(- 2-4)x2y +(-6-3)xy2 + (1+1)y3
= -x3 - 6x2y - 9xy2 + 2y3

(vi) Term to be subtracted = −11x2y2 + 7xy −6
Changing the sign of each term of the expression gives  11x2y2 -7xy +6.
(9x2y2 −6xy + 9)+(11x2y2 -7xy +6) = 9x2y2 −6xy + 9+11x2y2 -7xy +6
= (9+11)x2y2 (-7−6)xy + 15
= 20x2y2 −13xy +15

(vii) Term to be subtracted = −2a + b + 6d
Changing the sign of each term of the expression gives 2a-b-6d.
(5a − 2b -3c)+(2a-b-6d ) = 5a − 2b -3c +2a-b-6d
= (5+2)a+(− 2-1)b -3c -6d
= 7a − 3b-3c -6d

#### Question 6:

Change the sign of each term of the expression that is to be subtracted and then add.

(i) Term to be subtracted = 5a + 7b − 2c
Changing the sign of each term of the expression gives -5a -7b + 2c.
(3a − 7b + 4c)+(-5a -7b + 2c ) = 3a − 7b + 4c-5a -7b + 2c
= (3-5)a+( − 7-7)b + (4+2)c
= -2a − 14b + 6c

(ii) Term to be subtracted = a − 2b − 3c
Changing the sign of each term of the expression gives -a +2b + 3c.
(−2a + 5b − 4c)+(-a +2b + 3c ) = −2a + 5b − 4c-a +2b + 3c
= (−2-1)a + (5+2)b +(−4+3)c
= −3a + 7b − c

(iii) Term to be subtracted = 5x2 − 3xy + y2
Changing the sign of each term of the expression gives  -5x2 + 3xy - y2.

(7x2 − 2xy − 4y2)+(-5x2 + 3xy - y2) = 7x2 − 2xy − 4y2-5x2 + 3xy - y2
= (7-5)x2 +(−2+3)xy +(−4-1)y2
= 2x2 +xy − 5y2

(iv) Term to be subtracted = 6x3 − 7x2 + 5x − 3
Changing the sign of each term of the expression gives  -6x3 + 7x2 - 5x + 3.
(4 − 5x + 6x2 − 8x3)+(-6x3 + 7x2 - 5x + 3) = 4 − 5x + 6x2 − 8x3-6x3 + 7x2 - 5x + 3
= (-8-6)x3 +(6+7)x2 +(-5- 5)x + 7
= -14x3 + 13x2 - 10x + 7

(v) Term to be subtracted = x3 + 2x2y + 6xy2 − y3
Changing the sign of each term of the expression gives  -x3 - 2x2y - 6xy2 + y3.
(y3 − 3xy2 − 4x2y)+(-x3 - 2x2y - 6xy2 + y3) = y3 − 3xy2 − 4x2y-x3 - 2x2y - 6xy2 + y3
= -x3 +(- 2-4)x2y +(-6-3)xy2 + (1+1)y3
= -x3 - 6x2y - 9xy2 + 2y3

(vi) Term to be subtracted = −11x2y2 + 7xy −6
Changing the sign of each term of the expression gives  11x2y2 -7xy +6.
(9x2y2 −6xy + 9)+(11x2y2 -7xy +6) = 9x2y2 −6xy + 9+11x2y2 -7xy +6
= (9+11)x2y2 (-7−6)xy + 15
= 20x2y2 −13xy +15

(vii) Term to be subtracted = −2a + b + 6d
Changing the sign of each term of the expression gives 2a-b-6d.
(5a − 2b -3c)+(2a-b-6d ) = 5a − 2b -3c +2a-b-6d
= (5+2)a+(− 2-1)b -3c -6d
= 7a − 3b-3c -6d

(i) 2p3 − 3p2 + 4p − 5 − 6p3 + 2p2 − 8p − 2 + 6p + 8
Rearranging and collecting the like terms
= (2-6)p3 +(−3+2)p2 + (4-8+6)p − 5-2+8
= -4p3 −p2 +2p +1

(ii) 2x2 − xy + 6x − 4y + 5xy − 4x + 6x2 + 3y
Rearranging and collecting the like terms
= (2+6)x2 +(−1+5) xy + (6-4)x +(− 4+3)y
= 8x2 + 4xy + 2x − y

(iii) x4 − 6x3 + 2x − 7 + 7x3 − x + 5x2 + 2 − x4
Rearranging and collectingthe like terms
= (1-1)x4 +(− 6+7)x3 + 5x2 +(2-1)x-7+ 2
= 0 +  x3 + 5x2 +x-5
= x3 + 5x2 +x-5

#### Question 7:

(i) 2p3 − 3p2 + 4p − 5 − 6p3 + 2p2 − 8p − 2 + 6p + 8
Rearranging and collecting the like terms
= (2-6)p3 +(−3+2)p2 + (4-8+6)p − 5-2+8
= -4p3 −p2 +2p +1

(ii) 2x2 − xy + 6x − 4y + 5xy − 4x + 6x2 + 3y
Rearranging and collecting the like terms
= (2+6)x2 +(−1+5) xy + (6-4)x +(− 4+3)y
= 8x2 + 4xy + 2x − y

(iii) x4 − 6x3 + 2x − 7 + 7x3 − x + 5x2 + 2 − x4
Rearranging and collectingthe like terms
= (1-1)x4 +(− 6+7)x3 + 5x2 +(2-1)x-7+ 2
= 0 +  x3 + 5x2 +x-5
= x3 + 5x2 +x-5

(3x2 − 5x + 2) + (−5x2 − 8x + 6)
Rearranging and collecting the like terms:
(3-5)x2 +(− 5-8)x + 2 +6
= -2x2 − 13x + 8

Subtract 4x2 − 9x + 7 from -2x2 − 13x + 8.

Change the sign of each term of the expression that is to be subtracted and then add.

Term to be subtracted = 4x2 − 9x + 7
Changing the sign of each term of the expression gives -4x2 + 9x - 7.
( -2x2 − 13x + 8 )+(-4x2 + 9x - 7 )    = -2x2 − 13x + 8 -4x2 + 9x - 7
= ( -2-4)x2 +(−13+9)x + 8 -7
= -6x2 − 4x + 1

#### Question 8:

(3x2 − 5x + 2) + (−5x2 − 8x + 6)
Rearranging and collecting the like terms:
(3-5)x2 +(− 5-8)x + 2 +6
= -2x2 − 13x + 8

Subtract 4x2 − 9x + 7 from -2x2 − 13x + 8.

Change the sign of each term of the expression that is to be subtracted and then add.

Term to be subtracted = 4x2 − 9x + 7
Changing the sign of each term of the expression gives -4x2 + 9x - 7.
( -2x2 − 13x + 8 )+(-4x2 + 9x - 7 )    = -2x2 − 13x + 8 -4x2 + 9x - 7
= ( -2-4)x2 +(−13+9)x + 8 -7
= -6x2 − 4x + 1

A = 7x2 + 5xy − 9y2
B = −4x2 + xy + 5y2
C = 4y2 − 3x2 − 6xy

Substituting the values of A, B and C in A+B+C:
= (7x2 + 5xy − 9y2)+(−4x2 + xy + 5y2)+(4y2 − 3x2 − 6xy)
= 7x2 + 5xy − 9y2−4x2 + xy + 5y2+4y2 − 3x2 − 6xy

Rearranging and collecting the like terms:
(7-4-3)x2 + (5+1-6)xy +(−9+5+4)y2
= (0)x2 + (0)xy + (0)y2
= 0
$⇒\mathrm{A}+\mathrm{B}+\mathrm{C}=0$

#### Question 9:

A = 7x2 + 5xy − 9y2
B = −4x2 + xy + 5y2
C = 4y2 − 3x2 − 6xy

Substituting the values of A, B and C in A+B+C:
= (7x2 + 5xy − 9y2)+(−4x2 + xy + 5y2)+(4y2 − 3x2 − 6xy)
= 7x2 + 5xy − 9y2−4x2 + xy + 5y2+4y2 − 3x2 − 6xy

Rearranging and collecting the like terms:
(7-4-3)x2 + (5+1-6)xy +(−9+5+4)y2
= (0)x2 + (0)xy + (0)y2
= 0
$⇒\mathrm{A}+\mathrm{B}+\mathrm{C}=0$

Let the expression to be added be X.
(5x3 − 2x2 + 6x + 7)+X = (x3 + 3x2x + 1)
X = (x3 + 3x2x + 1) - (5x3 − 2x2 + 6x + 7)
Changing the sign of each term of the expression that is to be subtracted and then adding:
X = (x3 + 3x2x + 1) + (-5x3 + 2x2 - 6x - 7)
X = x3 + 3x2x + 1-5x3 + 2x2 - 6x - 7

Rearranging and collecting the like terms:
X = (1-5)x3 + (3+2)x2 +(−1-6) x + 1-7
X = -4x3 + 5x2 − 7x -6

So, -4x3 + 5x2 − 7x -6 must be added to 5x3 − 2x2 + 6x + 7 to get the sum as x3 + 3x2x + 1.

#### Question 10:

Let the expression to be added be X.
(5x3 − 2x2 + 6x + 7)+X = (x3 + 3x2x + 1)
X = (x3 + 3x2x + 1) - (5x3 − 2x2 + 6x + 7)
Changing the sign of each term of the expression that is to be subtracted and then adding:
X = (x3 + 3x2x + 1) + (-5x3 + 2x2 - 6x - 7)
X = x3 + 3x2x + 1-5x3 + 2x2 - 6x - 7

Rearranging and collecting the like terms:
X = (1-5)x3 + (3+2)x2 +(−1-6) x + 1-7
X = -4x3 + 5x2 − 7x -6

So, -4x3 + 5x2 − 7x -6 must be added to 5x3 − 2x2 + 6x + 7 to get the sum as x3 + 3x2x + 1.

P = a2 − b2 + 2ab
Q = a2 + 4b2 − 6ab
R = b2 + 6
S = a2 − 4ab
T = −2a2 + b2 − ab + a

Adding P, Q, R and S:
P+Q+R+S
= (a2 − b2 + 2ab)+(a2 + 4b2 − 6ab)+(b2 + 6)+(a2 − 4ab )
= a2 − b2 + 2ab+a2 + 4b2 − 6ab+b2 + 6+a2 − 4ab

Rearranging and collecting the like terms:
= (1+1+1)a2 +(−1+4+1) b2 + (2-6-4)ab+6
P+Q+R+S = 3a2 +4b2 - 8ab+6

To find P + Q + R + S − T, subtract T = (−2a2 + b2 − ab + a) from P+Q+R+S = (3a2 +4b2 - 8ab+6).

On changing the sign of each term of the expression that is to be subtracted and then adding:
Term to be subtracted = −2a2 + b2 − ab + a
Changing the sign of each term of the expression gives 2a2 - b2 + ab - a.
(3a2 +4b2 - 8ab+6)+(2a2 - b2 + ab - a) = 3a2 +4b2 - 8ab+6+2a2 - b2 + ab - a
= (3+2)a2 +(4-1) b2 +(-8+1) ab - a+6

P + Q + R + S − T = 5a2 +3b2 -7 ab - a+6

#### Question 11:

P = a2 − b2 + 2ab
Q = a2 + 4b2 − 6ab
R = b2 + 6
S = a2 − 4ab
T = −2a2 + b2 − ab + a

Adding P, Q, R and S:
P+Q+R+S
= (a2 − b2 + 2ab)+(a2 + 4b2 − 6ab)+(b2 + 6)+(a2 − 4ab )
= a2 − b2 + 2ab+a2 + 4b2 − 6ab+b2 + 6+a2 − 4ab

Rearranging and collecting the like terms:
= (1+1+1)a2 +(−1+4+1) b2 + (2-6-4)ab+6
P+Q+R+S = 3a2 +4b2 - 8ab+6

To find P + Q + R + S − T, subtract T = (−2a2 + b2 − ab + a) from P+Q+R+S = (3a2 +4b2 - 8ab+6).

On changing the sign of each term of the expression that is to be subtracted and then adding:
Term to be subtracted = −2a2 + b2 − ab + a
Changing the sign of each term of the expression gives 2a2 - b2 + ab - a.
(3a2 +4b2 - 8ab+6)+(2a2 - b2 + ab - a) = 3a2 +4b2 - 8ab+6+2a2 - b2 + ab - a
= (3+2)a2 +(4-1) b2 +(-8+1) ab - a+6

P + Q + R + S − T = 5a2 +3b2 -7 ab - a+6

Let the expression to be subtracted be X.
(a3 − 4a2 + 5a − 6)-X = (a2 − 2a + 1)
X = (a3 − 4a2 + 5a − 6)- (a2 − 2a + 1)
Since '-' sign precedes the parenthesis, we remove it and change the sign of each term within the parenthesis.
X = a3 − 4a2 + 5a − 6- a2 + 2a - 1
Rearranging and collecting the like terms:
X = a3 +(− 4-1)a2 + (5+2)a − 6 - 1
X = a3 −5a2 + 7a − 7
So, a3 −5a2 + 7a − 7 must be subtracted from a3 − 4a2 + 5a − 6 to obtain a2 − 2a + 1.

#### Question 12:

Let the expression to be subtracted be X.
(a3 − 4a2 + 5a − 6)-X = (a2 − 2a + 1)
X = (a3 − 4a2 + 5a − 6)- (a2 − 2a + 1)
Since '-' sign precedes the parenthesis, we remove it and change the sign of each term within the parenthesis.
X = a3 − 4a2 + 5a − 6- a2 + 2a - 1
Rearranging and collecting the like terms:
X = a3 +(− 4-1)a2 + (5+2)a − 6 - 1
X = a3 −5a2 + 7a − 7
So, a3 −5a2 + 7a − 7 must be subtracted from a3 − 4a2 + 5a − 6 to obtain a2 − 2a + 1.

To calculate how much is a + 2b − 3c greater than 2a − 3b + c, we have to subtract 2a − 3b + c from a + 2b − 3c.

Change the sign of each term of the expression that is to be subtracted and then add.

Term to be subtracted = 2a − 3b + c
Changing the sign of each term of the expression gives -2a + 3b - c.
(a + 2b − 3c )+(-2a + 3b - c )
= a + 2b − 3c -2a + 3b - c
= (1-2)a + (2+3)b +(− 3-1)c
= -a + 5b − 4c

#### Question 13:

To calculate how much is a + 2b − 3c greater than 2a − 3b + c, we have to subtract 2a − 3b + c from a + 2b − 3c.

Change the sign of each term of the expression that is to be subtracted and then add.

Term to be subtracted = 2a − 3b + c
Changing the sign of each term of the expression gives -2a + 3b - c.
(a + 2b − 3c )+(-2a + 3b - c )
= a + 2b − 3c -2a + 3b - c
= (1-2)a + (2+3)b +(− 3-1)c
= -a + 5b − 4c

To calculate how much less than x − 2y + 3z is 2x − 4y − z, we have to subtract 2x − 4y − z from x − 2y + 3z.

Change the sign of each term of the expression that is to be subtracted and then add.

Term to be subtracted = 2x − 4y − z
Changing the sign of each term of the expression gives -2x + 4y + z.
(x − 2y + 3z)+(-2x + 4y + z )
= x − 2y + 3z-2x + 4y + z
= (1-2)x +(−2+4)y + (3+1)z
= -x + 2y + 4z

#### Question 14:

To calculate how much less than x − 2y + 3z is 2x − 4y − z, we have to subtract 2x − 4y − z from x − 2y + 3z.

Change the sign of each term of the expression that is to be subtracted and then add.

Term to be subtracted = 2x − 4y − z
Changing the sign of each term of the expression gives -2x + 4y + z.
(x − 2y + 3z)+(-2x + 4y + z )
= x − 2y + 3z-2x + 4y + z
= (1-2)x +(−2+4)y + (3+1)z
= -x + 2y + 4z

To calculate how much does 3x2 − 5x + 6 exceed x3 − x2 + 4x − 1, we have to subtract x3 − x2 + 4x − 1 from 3x2 − 5x + 6.
Change the sign of each term of the expression that is to be subtracted and then add.

Term to be subtracted = x3 − x2 + 4x − 1
Changing the sign of each term of the expression gives -x3 + x2 - 4x + 1.
(3x2 − 5x + 6)+(-x3 + x2 - 4x + 1 )
= 3x2 − 5x + 6-x3 + x2 - 4x + 1
= -x3 + (3+1)x2 +(-5-4)x+6 + 1
= -x3 +4 x2 - 9x + 7

#### Question 15:

To calculate how much does 3x2 − 5x + 6 exceed x3 − x2 + 4x − 1, we have to subtract x3 − x2 + 4x − 1 from 3x2 − 5x + 6.
Change the sign of each term of the expression that is to be subtracted and then add.

Term to be subtracted = x3 − x2 + 4x − 1
Changing the sign of each term of the expression gives -x3 + x2 - 4x + 1.
(3x2 − 5x + 6)+(-x3 + x2 - 4x + 1 )
= 3x2 − 5x + 6-x3 + x2 - 4x + 1
= -x3 + (3+1)x2 +(-5-4)x+6 + 1
= -x3 +4 x2 - 9x + 7

Add 5x − 4y + 6z and −8x + y − 2z.

(5x − 4y + 6z )+(−8x + y − 2z)
= 5x − 4y + 6z −8x + y − 2z
= (5-8)x +(−4+1)y + (6-2)z
= -3x − 3y + 4z

Adding 12x − y + 3z and −3x + 5y − 8z:
(12x − y + 3z )+(−3x + 5y − 8z)
= 12x − y + 3z −3x + 5y − 8z
= (12-3)x +(−1+5)y + (3-8)z
= 9x +4y -5z

Subtract -3x − 3y + 4z from 9x +4y -5z.
Change the sign of each term of the expression that is to be subtracted and then add.

Term to be subtracted = -3x − 3y + 4z
Changing the sign of each term of the expression gives 3x + 3y - 4z.
(9x +4y -5z)+(3x + 3y - 4z )
= 9x +4y -5z+3x + 3y - 4z
= (9+3)x +(4+3)y + (-5-4)z
= 12x +7y -9z

#### Question 16:

Add 5x − 4y + 6z and −8x + y − 2z.

(5x − 4y + 6z )+(−8x + y − 2z)
= 5x − 4y + 6z −8x + y − 2z
= (5-8)x +(−4+1)y + (6-2)z
= -3x − 3y + 4z

Adding 12x − y + 3z and −3x + 5y − 8z:
(12x − y + 3z )+(−3x + 5y − 8z)
= 12x − y + 3z −3x + 5y − 8z
= (12-3)x +(−1+5)y + (3-8)z
= 9x +4y -5z

Subtract -3x − 3y + 4z from 9x +4y -5z.
Change the sign of each term of the expression that is to be subtracted and then add.

Term to be subtracted = -3x − 3y + 4z
Changing the sign of each term of the expression gives 3x + 3y - 4z.
(9x +4y -5z)+(3x + 3y - 4z )
= 9x +4y -5z+3x + 3y - 4z
= (9+3)x +(4+3)y + (-5-4)z
= 12x +7y -9z

To calculate how much is 2x − 3y + 4z greater than 2x + 5y − 6z + 2, we have to subtract 2x + 5y − 6z + 2 from 2x − 3y + 4z.
Change the sign of each term of the expression that is to be subtracted and then add.

Term to be subtracted = 2x + 5y − 6z + 2
Changing the sign of each term of the expression gives -2x - 5y + 6z - 2.
(2x − 3y + 4z )+(-2x - 5y + 6z - 2 )
= 2x − 3y + 4z-2x - 5y + 6z - 2
= (2-2)x + (-3-5)y +(4+6)z-2
= 0-8y+10z-2
= -8y+10z-2

#### Question 17:

To calculate how much is 2x − 3y + 4z greater than 2x + 5y − 6z + 2, we have to subtract 2x + 5y − 6z + 2 from 2x − 3y + 4z.
Change the sign of each term of the expression that is to be subtracted and then add.

Term to be subtracted = 2x + 5y − 6z + 2
Changing the sign of each term of the expression gives -2x - 5y + 6z - 2.
(2x − 3y + 4z )+(-2x - 5y + 6z - 2 )
= 2x − 3y + 4z-2x - 5y + 6z - 2
= (2-2)x + (-3-5)y +(4+6)z-2
= 0-8y+10z-2
= -8y+10z-2

To calculate how much does 1 exceed 2x-3y-4, we have to subtract 2x-3y-4 from 1.
Change the sign of each term of the expression to be subtracted and then add.

Term to be subtracted = 2x-3y-4
Changing the sign of each term of the expression gives -2x+3y+4.
(1)+(-2x+3y+4 )
= 1-2x+3y+4
= 5-2x+3y

#### Question 1:

To calculate how much does 1 exceed 2x-3y-4, we have to subtract 2x-3y-4 from 1.
Change the sign of each term of the expression to be subtracted and then add.

Term to be subtracted = 2x-3y-4
Changing the sign of each term of the expression gives -2x+3y+4.
(1)+(-2x+3y+4 )
= 1-2x+3y+4
= 5-2x+3y

a - (b - 2a)
Here,  '-' sign precedes the parenthesis. So, we will remove it and change the sign of each term within the parenthesis.
=a - b + 2a
=3a - b

#### Question 2:

a - (b - 2a)
Here,  '-' sign precedes the parenthesis. So, we will remove it and change the sign of each term within the parenthesis.
=a - b + 2a
=3a - b

4x − (3y − x + 2z)
Here, '−' sign precedes the parenthesis. So, we will remove it and change the sign of each term within the parenthesis.
= 4x − 3y + x − 2z
= 5x − 3y − 2z

#### Question 3:

4x − (3y − x + 2z)
Here, '−' sign precedes the parenthesis. So, we will remove it and change the sign of each term within the parenthesis.
= 4x − 3y + x − 2z
= 5x − 3y − 2z

(a2 + b2 + 2ab) − (a2 + b2 − 2ab)
Here, '−' sign precedes the second parenthesis. So, we will remove it and change the sign of each term within the parenthesis.
a2 + b2 + 2ab − a2 − b2 +2ab

Rearranging and collecting the like terms:
a2 − a2 +b2 − b2 + 2ab + 2ab
=(1 − 1)a2 + (1− 1)b2 + (2 + 2)ab
=0 + 0 + 4ab
= 4ab

#### Question 4:

(a2 + b2 + 2ab) − (a2 + b2 − 2ab)
Here, '−' sign precedes the second parenthesis. So, we will remove it and change the sign of each term within the parenthesis.
a2 + b2 + 2ab − a2 − b2 +2ab

Rearranging and collecting the like terms:
a2 − a2 +b2 − b2 + 2ab + 2ab
=(1 − 1)a2 + (1− 1)b2 + (2 + 2)ab
=0 + 0 + 4ab
= 4ab

−3(a + b) + 4(2a − 3b) − (2a − b)
Here, '−' sign precedes the first and the third parenthesis. So, we will remove them and change the sign of each term within the two parenthesis.
= −3a − 3b + (4$×$2a )−(4$×$3b) − 2a + b
= − 3a − 3b + 8a − 12b − 2a + b

Rearranging and collecting the like terms:
−3a + 8a − 2a − 3b − 12b + b
= (−3 + 8 − 2)a + (−3 − 12 + 1)b
= 3a −14b

#### Question 5:

−3(a + b) + 4(2a − 3b) − (2a − b)
Here, '−' sign precedes the first and the third parenthesis. So, we will remove them and change the sign of each term within the two parenthesis.
= −3a − 3b + (4$×$2a )−(4$×$3b) − 2a + b
= − 3a − 3b + 8a − 12b − 2a + b

Rearranging and collecting the like terms:
−3a + 8a − 2a − 3b − 12b + b
= (−3 + 8 − 2)a + (−3 − 12 + 1)b
= 3a −14b

−4x2 + {(2x2 − 3) − (4 − 3x2)}

We will first remove the innermost grouping symbol (  ) and then {  }.

∴ −4x2 + {(2x2 − 3) − (4 − 3x2)}
= −4x2 + {2x2 − 3 − 4 + 3x2}
= −4x2 + {5x2 − 7}
= −4x2 + 5x2 − 7
= x2 − 7

#### Question 6:

−4x2 + {(2x2 − 3) − (4 − 3x2)}

We will first remove the innermost grouping symbol (  ) and then {  }.

∴ −4x2 + {(2x2 − 3) − (4 − 3x2)}
= −4x2 + {2x2 − 3 − 4 + 3x2}
= −4x2 + {5x2 − 7}
= −4x2 + 5x2 − 7
= x2 − 7

−2(x2y2 + xy) −3(x2 + y2xy)
Here a '−' sign precedes both the parenthesis. So, we will remove them and change the sign of each term within the two parenthesis.
= −2x2 +2y2 − 2xy −3x2 − 3y2 + 3xy
= (−2 − 3)x2 +(2 − 3)y2 + (− 2 + 3)xy
= −5x2y2 + xy

#### Question 7:

−2(x2y2 + xy) −3(x2 + y2xy)
Here a '−' sign precedes both the parenthesis. So, we will remove them and change the sign of each term within the two parenthesis.
= −2x2 +2y2 − 2xy −3x2 − 3y2 + 3xy
= (−2 − 3)x2 +(2 − 3)y2 + (− 2 + 3)xy
= −5x2y2 + xy

a − [2b − {3a − (2b − 3c)}]
We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ a − [2b − {3a − (2b − 3c)}]
= a − [2b − {3a − 2b + 3c}]
= a − [2b − 3a + 2b − 3c]
= a − [4b − 3a − 3c]
= a − 4b + 3a + 3c
= 4a − 4b + 3c

#### Question 8:

a − [2b − {3a − (2b − 3c)}]
We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ a − [2b − {3a − (2b − 3c)}]
= a − [2b − {3a − 2b + 3c}]
= a − [2b − 3a + 2b − 3c]
= a − [4b − 3a − 3c]
= a − 4b + 3a + 3c
= 4a − 4b + 3c

−x + [5y − {x − (5y − 2x)}]
We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ −x + [5y − {x − (5y − 2x)}]
= −x + [5y − {x − 5y + 2x}]
= −x + [5y − {3x − 5y}]
= −x + [5y − 3x + 5y]
= −x + [10y − 3x]
= −x + 10y − 3x
=  − 4x + 10y

#### Question 9:

−x + [5y − {x − (5y − 2x)}]
We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ −x + [5y − {x − (5y − 2x)}]
= −x + [5y − {x − 5y + 2x}]
= −x + [5y − {3x − 5y}]
= −x + [5y − 3x + 5y]
= −x + [10y − 3x]
= −x + 10y − 3x
=  − 4x + 10y

86 − [15x − 7(6x − 9) −2{10x − 5(2 − 3x)}]
We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ 86 − [15x − 7(6x − 9) −2{10x − 5(2 − 3x)}]
= 86 − [15x − 42x + 63 −2{10x − 10 + 15x}]
= 86 − [15x − 42x + 63 −2{25x − 10}]
= 86 − [15x − 42x + 63 −50x + 20]
= 86 − [− 77x + 83 ]
= 86 + 77x − 83
= 77x + 3

#### Question 10:

86 − [15x − 7(6x − 9) −2{10x − 5(2 − 3x)}]
We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ 86 − [15x − 7(6x − 9) −2{10x − 5(2 − 3x)}]
= 86 − [15x − 42x + 63 −2{10x − 10 + 15x}]
= 86 − [15x − 42x + 63 −2{25x − 10}]
= 86 − [15x − 42x + 63 −50x + 20]
= 86 − [− 77x + 83 ]
= 86 + 77x − 83
= 77x + 3

12x − [3x3 + 5x2 − {7x2 − (4 − 3x − x3) + 6x3} − 3x]
We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ 12x − [3x3 + 5x2 − {7x2 − (4 − 3x − x3) + 6x3} − 3x]
= 12x − [3x3 + 5x2 − {7x2 − 4 + 3x + x3+ 6x3} − 3x]
= 12x − [3x3 + 5x2 − {7x2 − 4 + 3x + 7x3} − 3x]
= 12x − [3x3 + 5x2 − 7x2 + 4 − 3x − 7x3 − 3x]
= 12x − [ − 2x2 + 4 − 4x3 − 6x]
= 12x + 2x2 − 4 + 4x3 + 6x
= 4x3 + 2x2 +18x-4

#### Question 11:

12x − [3x3 + 5x2 − {7x2 − (4 − 3x − x3) + 6x3} − 3x]
We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ 12x − [3x3 + 5x2 − {7x2 − (4 − 3x − x3) + 6x3} − 3x]
= 12x − [3x3 + 5x2 − {7x2 − 4 + 3x + x3+ 6x3} − 3x]
= 12x − [3x3 + 5x2 − {7x2 − 4 + 3x + 7x3} − 3x]
= 12x − [3x3 + 5x2 − 7x2 + 4 − 3x − 7x3 − 3x]
= 12x − [ − 2x2 + 4 − 4x3 − 6x]
= 12x + 2x2 − 4 + 4x3 + 6x
= 4x3 + 2x2 +18x-4

5a − [a2 − {2a(1 − a + 4a2) − 3a(a2 − 5a − 3)}] −8a
We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ 5a − [a2 − {2a(1 − a + 4a2) − 3a(a2 − 5a − 3)}] −8a
= 5a − [a2 − {2a − 2a2 + 8a3 − 3a3 + 15a2 + 9a}] −8a
= 5a − [a2 − {5a3 + 13a2 + 11a}] − 8a
= 5a − [a2 − 5a3 − 13a2 −11a] − 8a
= 5a − [ − 5a3 − 12a2 − 11a] − 8a
= 5a + 5a3 + 12a2 + 11a − 8a
= 5a3 + 12a2 + 8a

#### Question 12:

5a − [a2 − {2a(1 − a + 4a2) − 3a(a2 − 5a − 3)}] −8a
We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ 5a − [a2 − {2a(1 − a + 4a2) − 3a(a2 − 5a − 3)}] −8a
= 5a − [a2 − {2a − 2a2 + 8a3 − 3a3 + 15a2 + 9a}] −8a
= 5a − [a2 − {5a3 + 13a2 + 11a}] − 8a
= 5a − [a2 − 5a3 − 13a2 −11a] − 8a
= 5a − [ − 5a3 − 12a2 − 11a] − 8a
= 5a + 5a3 + 12a2 + 11a − 8a
= 5a3 + 12a2 + 8a

3 − [x − {2y − (5x + y − 3) + 2x2} − (x2 − 3y)]

We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ 3 − [x − {2y − (5x + y − 3) + 2x2} − (x2 − 3y)]
= 3 − [x − {2y − 5x - y + 3 + 2x2} − x2 + 3y]
= 3 − [x − {y − 5x + 3 + 2x2} − x2 + 3y]
= 3 − [x − y + 5x − 3 − 2x2 − x2 + 3y]
= 3 − [ 6x − 3 − 3x2 + 2y]
= 3 − 6x + 3 + 3x2 − 2y
= 3x2 − 2y − 6x+6

#### Question 13:

3 − [x − {2y − (5x + y − 3) + 2x2} − (x2 − 3y)]

We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ 3 − [x − {2y − (5x + y − 3) + 2x2} − (x2 − 3y)]
= 3 − [x − {2y − 5x - y + 3 + 2x2} − x2 + 3y]
= 3 − [x − {y − 5x + 3 + 2x2} − x2 + 3y]
= 3 − [x − y + 5x − 3 − 2x2 − x2 + 3y]
= 3 − [ 6x − 3 − 3x2 + 2y]
= 3 − 6x + 3 + 3x2 − 2y
= 3x2 − 2y − 6x+6

xy − [yzzx − {yx − (3yxz) − (xyzy)}]

We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ xy − [yzzx − {yx − (3yxz) − (xyzy)}]
= xy − [yzzx − {yx − 3y + xzxy + zy}]
= xy − [yzzx − {− 3y + xz + zy}]  $\left(\because \mathrm{xy}=\mathrm{yx}\right)$
= xy − [yzzx + 3y - xz - zy
= xy − [− 2zx + 3y
= xy + 2zx − 3y

#### Question 14:

xy − [yzzx − {yx − (3yxz) − (xyzy)}]

We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ xy − [yzzx − {yx − (3yxz) − (xyzy)}]
= xy − [yzzx − {yx − 3y + xzxy + zy}]
= xy − [yzzx − {− 3y + xz + zy}]  $\left(\because \mathrm{xy}=\mathrm{yx}\right)$
= xy − [yzzx + 3y - xz - zy
= xy − [− 2zx + 3y
= xy + 2zx − 3y

2a − 3b − [3a − 2b − {a − c − (a − 2b)}]
We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ 2a − 3b − [3a − 2b − {a − c − (a − 2b)}]
= 2a − 3b − [3a − 2b − {a − c − a + 2b}]
= 2a − 3b − [3a − 2b − {− c  + 2b}]
= 2a − 3b − [3a − 2b + c  − 2b]
= 2a − 3b − [3a − 4b + c ]
= 2a − 3b − 3a + 4b − c
= − a + b − c

#### Question 15:

2a − 3b − [3a − 2b − {a − c − (a − 2b)}]
We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ 2a − 3b − [3a − 2b − {a − c − (a − 2b)}]
= 2a − 3b − [3a − 2b − {a − c − a + 2b}]
= 2a − 3b − [3a − 2b − {− c  + 2b}]
= 2a − 3b − [3a − 2b + c  − 2b]
= 2a − 3b − [3a − 4b + c ]
= 2a − 3b − 3a + 4b − c
= − a + b − c

-a − [a + {a + b − 2a − (a − 2b)} − b]
We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ −a − [a + {a + b − 2a − (a − 2b)} − b]
= −a − [a + {a + b − 2a − a + 2b} − b]
= −a − [a + {3b − 2a } − b]
= −a − [a + 3b − 2a  − b]
= −a − [2b − a ]
= −a − 2b + a
= −2b

#### Question 16:

-a − [a + {a + b − 2a − (a − 2b)} − b]
We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ −a − [a + {a + b − 2a − (a − 2b)} − b]
= −a − [a + {a + b − 2a − a + 2b} − b]
= −a − [a + {3b − 2a } − b]
= −a − [a + 3b − 2a  − b]
= −a − [2b − a ]
= −a − 2b + a
= −2b

$2a-\left[4b-\left\{4a-\left(3b-\overline{2a+2b}\right)\right\}\right]$
We will first remove the innermost grouping symbol bar bracket. Next, we will remove (  ), followed by {  } and then [   ].

$2a-\left[4b-\left\{4a-\left(3b-\overline{2a+2b}\right)\right\}\right]$
= 2a-[4b-{4a-(3b-2a-2b)}]
= 2a-[4b-{4a-(b-2a)}]
= 2a-[4b-{4a-b+2a}]
=2a-[4b-{6a-b}]
= 2a-[4b-6a+b]
= 2a-[5b-6a]
= 2a-5b+6a
= 8a-5b

#### Question 17:

$2a-\left[4b-\left\{4a-\left(3b-\overline{2a+2b}\right)\right\}\right]$
We will first remove the innermost grouping symbol bar bracket. Next, we will remove (  ), followed by {  } and then [   ].

$2a-\left[4b-\left\{4a-\left(3b-\overline{2a+2b}\right)\right\}\right]$
= 2a-[4b-{4a-(3b-2a-2b)}]
= 2a-[4b-{4a-(b-2a)}]
= 2a-[4b-{4a-b+2a}]
=2a-[4b-{6a-b}]
= 2a-[4b-6a+b]
= 2a-[5b-6a]
= 2a-5b+6a
= 8a-5b

5x − [4y − {7x − (3z − 2y) + 4z − 3(x + 3y − 2z)}]
We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ 5x − [4y − {7x − (3z − 2y) + 4z − 3(x + 3y − 2z)}]
= 5x − [4y − {7x − 3z + 2y + 4z − 3x − 9y + 6z}]
= 5x − [4y − {4x + 7z − 7y}]
= 5x − [4y − 4x − 7z + 7y]
= 5x − [11y − 4x − 7z ]
= 5x − 11y + 4x + 7z
= 9x − 11y + 7z

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