Linear Equations - Exercise 14(A)

Solve the following equation and check your answer:

5(z + 4) = 35

Answer

We have:

5(z + 4) = 35

⇒ z + 4 = 7 $\quad$[Dividing both sides by 5]

⇒ z = 7 - 4 $\quad$[Transposing +4 to RHS]

∴ z = 3

Check:

LHS = 5(z + 4)

= 5(3 + 4)

= 5 × 7

= 35

RHS = 35

Hence, LHS = RHS.

Solve the following equation and check your answer:

$2\Big(y -\dfrac{5}{2}\Big) = 0.3$

Answer

We have:

$\phantom{=} 2\Big(y -\dfrac{5}{2}\Big) = 0.3 \\[1em] \Rightarrow 2y - 2 \times \dfrac{5}{2} = 0.3 \quad \text{[Removing brackets]} \\[1em] \Rightarrow 2y - 5 = 0.3 \\[1em] \Rightarrow 2y = 0.3 + 5 \quad\text{[Transposing -5 to RHS]} \\[1em] \Rightarrow 2y = 5.3 \\[1em] \Rightarrow y = \dfrac{5.3}{2}$

∴ y = 2.65

Check:

LHS = $2\left(y - \dfrac{5}{2}\right)$

= 2(2.65 - 2.5)

= 2(0.15) = 0.3

RHS = 0.3

Hence, LHS = RHS.

Solve the following equation and check your answer:

$\dfrac{5}{8}x - 6 = 9$

Answer

We have:

$\phantom{=} \Rightarrow \dfrac{5}{8}x - 6 = 9 \\[1em] \Rightarrow \dfrac{5}{8}x = 9 + 6 \quad\text{[Transposing -6 to RHS]} \\[1em] \Rightarrow \dfrac{5}{8}x = 15 \\[1em] \Rightarrow x = 15 \times \dfrac{8}{5} \\[1em] \Rightarrow x = 3 \times \dfrac{8}{1}$

∴ x = 24

Check:

LHS = $\dfrac{5}{8}x$ - 6

= $\dfrac{5}{8}$ × 24 - 6

= 15 - 6

= 9

RHS = 9

Hence, LHS = RHS.

Solve the following equation and check your answer:

$3x - \dfrac{5}{3} = x - 3$

Answer

We have:

$\phantom{=} 3x - \dfrac{5}{3} = x - 3 \\[1em] \Rightarrow 3x - x = \dfrac{5}{3} - 3 \quad \text{[Transposing } -\dfrac{5}{3} \text{ to RHS and +x to LHS]} \\[1em] \Rightarrow 2x = \dfrac{5 - 9}{3} \\[1em] \Rightarrow 2x = -\dfrac{4}{3} \\[1em] \Rightarrow x = -\dfrac{4}{3 \times 2} \\[1em] \Rightarrow x = -\dfrac{2}{3 \times 1}$

∴ x = $-\dfrac{2}{3}$

Check:

$\text{LHS} = 3x - \dfrac{5}{3} \\[1em] \phantom{\text{LHS}} = 3\left(-\dfrac{2}{3}\right) - \dfrac{5}{3} \\[1em] \phantom{\text{LHS}} = -2 - \dfrac{5}{3} \\[1em] \phantom{\text{LHS}} = -\dfrac{11}{3} \\[2em] \text{RHS} = x - 3 \\[1em] \phantom{\text{RHS}} = -\dfrac{2}{3} - 3 \\[1em] \phantom{\text{RHS}} = -\dfrac{11}{3}$

Hence, LHS = RHS.

Solve the following equation and check your answer:

0.6x - 1.9 = 0.2x + 0.5

Answer

We have:

0.6x - 1.9 = 0.2x + 0.5

⇒ 0.6x - 0.2x = 0.5 + 1.9 $\quad$[Transposing -1.9 to RHS and +0.2x to LHS]

⇒ 0.4x = 2.4

⇒ $x = \dfrac{2.4}{0.4}$

∴ x = 6

Check:

LHS = 0.6x - 1.9

= 0.6 × 6 - 1.9

= 3.6 - 1.9

= 1.7

RHS = 0.2x + 0.5

= 0.2 × 6 + 0.5

= 1.2 + 0.5

= 1.7

Hence, LHS = RHS.

Solve the following equation and check your answer:

2(3y - 2) - 4(2y - 5) = 9

Answer

We have:

2(3y - 2) - 4(2y - 5) = 9

⇒ 6y - 4 - 8y + 20 = 9 $\quad$[Removing brackets]

⇒ -2y + 16 = 9

⇒ -2y = 9 - 16 $\quad$[Transposing +16 to RHS]

⇒ -2y = -7

⇒ y = $\dfrac{-7}{-2}$

∴ y = $3\dfrac{1}{2}$

Check:

LHS = 2(3y - 2) - 4(2y - 5)

= $2\left(3 × \dfrac{7}{2} - 2\right)$ - $4\left(2 × \dfrac{7}{2} - 5\right)$

= $2\left(\dfrac{17}{2}\right)$ - 4(2)

= 17 - 8

= 9

RHS = 9

Hence, LHS = RHS.

Solve the following equation and check your answer:

5(3 - x) + 1 = 3(x + 4)

Answer

We have:

5(3 - x) + 1 = 3(x + 4)

⇒ 15 - 5x + 1 = 3x + 12 $\quad$[Removing brackets]

⇒ 16 - 5x = 3x + 12

⇒ 16 - 12 = 3x + 5x $\quad$[Transposing -5x to RHS and +12 to LHS]

⇒ 4 = 8x

⇒ x = $\dfrac{4}{8}$

∴ x = $\dfrac{1}{2}$

Check:

$\text{LHS} = 5(3 - x) + 1 \\[1em] \phantom{\text{LHS}} = 5\left(3 - \dfrac{1}{2}\right) + 1 \\[1em] \phantom{\text{LHS}} = 5\left(\dfrac{5}{2}\right) + 1 \\[1em] \phantom{\text{LHS}} = \dfrac{25}{2} + 1 \\[1em] \phantom{\text{LHS}} = \dfrac{27}{2} \\[2em] \text{RHS} = 3(x + 4) \\[1em] \phantom{\text{RHS}} = 3\left(\dfrac{1}{2} + 4\right) \\[1em] \phantom{\text{RHS}} = 3\left(\dfrac{9}{2}\right) \\[1em] \phantom{\text{RHS}} = \dfrac{27}{2}$

Hence, LHS = RHS.

Solve the following equation and check your answer:

7 - 2(5 - 3x) = 4(x - 3) + 5

Answer

We have:

7 - 2(5 - 3x) = 4(x - 3) + 5

⇒ 7 - 10 + 6x = 4x - 12 + 5 $\quad$[Removing brackets]

⇒ -3 + 6x = 4x - 7

⇒ 6x - 4x = -7 + 3 $\quad$[Transposing -3 to RHS and +4x to LHS]

⇒ 2x = -4

⇒ x = $\dfrac{-4}{2}$

⇒ x = $\dfrac{-2}{1}$

∴ x = -2

Check:

LHS = 7 - 2(5 - 3x)

= 7 - 2[5 - 3(-2)]

= 7 - 2(11)

= -15

RHS = 4(x - 3) + 5

= 4(-2 - 3) + 5

= -20 + 5

= -15

Hence, LHS = RHS.

Solve the following equation and check your answer:

6(3x + 2) - 5(6x - 1) = 3(x - 8) - 5(7x - 6) + 9x

Answer

We have:

6(3x + 2) - 5(6x - 1) = 3(x - 8) - 5(7x - 6) + 9x

⇒ 18x + 12 - 30x + 5 = 3x - 24 - 35x + 30 + 9x $\quad$[Removing brackets]

⇒ -12x + 17 = -23x + 6

⇒ -12x + 23x = 6 - 17 $\quad$[Transposing +17 to RHS and -23x to LHS]

⇒ 11x = -11

⇒ x = $\dfrac{-11}{11}$

⇒ x = $\dfrac{-1}{1}$

∴ x = -1

Check:

LHS = 6(3x + 2) - 5(6x - 1)

= 6(-1) - 5(-7)

= -6 + 35

= 29

RHS = 3(x - 8) - 5(7x - 6) + 9x

= 3(-9) - 5(-13) - 9

= -27 + 65 - 9

= 29

Hence, LHS = RHS.

Solve the following equation and check your answer:

p - (2p + 5) - 5(1 - 2p) = 2(3 + 4p) - 3(p - 4)

Answer

We have:

p - (2p + 5) - 5(1 - 2p) = 2(3 + 4p) - 3(p - 4)

⇒ p - 2p - 5 - 5 + 10p = 6 + 8p - 3p + 12 $\quad$[Removing brackets]

⇒ 9p - 10 = 5p + 18

⇒ 9p - 5p = 18 + 10 $\quad$[Transposing -10 to RHS and +5p to LHS]

⇒ 4p = 28

⇒ p = $\dfrac{28}{4}$

⇒ p = $\dfrac{7}{1}$

∴ p = 7

Check:

LHS = p - (2p + 5) - 5(1 - 2p)

= 7 - 19 - 5(-13)

= -12 + 65

= 53

RHS = 2(3 + 4p) - 3(p - 4)

= 2(31) - 3(3)

= 62 - 9

= 53

Hence, LHS = RHS.

Solve the following equation and check your answer:

$\dfrac{2x + 3}{3 +x} = \dfrac{3}{2}$

Answer

We have:

$\phantom{=} \dfrac{2x + 3}{3 +x} = \dfrac{3}{2} \\[1em] \Rightarrow 2(2x + 3) = 3(3 + x) \\[1em] \text{[By cross multiplication]} \\[1em] \Rightarrow 4x + 6 = 9 + 3x \\[1em] \Rightarrow 4x - 3x = 9 - 6 \\[1em] \quad \text{[Transposing +6 to RHS and +3x to LHS]}$

∴ x = 3

Check:

LHS = $\dfrac{2x + 3}{3 + x}$

= $\dfrac{2(3) + 3}{3 + 3}$

= $\dfrac{9}{6}$

= $\dfrac{3}{2}$

RHS = $\dfrac{3}{2}$

Hence, LHS = RHS.

Solve the following equation and check your answer:

$\dfrac{8 - 3x}{5x + 31} = \dfrac{2}{3}$

Answer

We have:

$\phantom{=} \dfrac{8 - 3x}{5x + 31} = \dfrac{2}{3} \\[1em] \Rightarrow 3(8 - 3x) = 2(5x + 31) \\[1em] \quad \text{[By cross multiplication]} \\[1em] \Rightarrow 24 - 9x = 10x + 62 \\[1em] \Rightarrow 24 - 62 = 10x + 9x \\[1em] \quad \text{[Transposing -9x to RHS and +62 to LHS]} \\[1em] \Rightarrow -38 = 19x \\[1em] \Rightarrow x = \dfrac{-38}{19} \\[1em]$

∴ x = -2

Check:

LHS = $\dfrac{8 - 3x}{5x + 31}$

= $\dfrac{8 - 3(-2)}{5(-2) + 31}$

= $\dfrac{14}{21}$

= $\dfrac{2}{3}$

RHS = $\dfrac{2}{3}$

Hence, LHS = RHS.

Solve the following equation and check your answer:

$\dfrac{x}{3}+\dfrac{x}{4} = 14$

Answer

We have:

$\phantom{=} \dfrac{x}{3}+\dfrac{x}{4} = 14 \\[1em] \Rightarrow \dfrac{4x + 3x}{12} = 14 \\[1em] \Rightarrow 4x + 3x = 14 \times 12 \\[1em] \Rightarrow 7x = 168 \\[1em] \Rightarrow x = \dfrac{168}{7}$

∴ x = 24

Check:

LHS = $\dfrac{x}{3} + \dfrac{x}{4}$

= $\dfrac{24}{3} + \dfrac{24}{4}$

= 8 + 6 = 14

RHS = 14

Hence, LHS = RHS.

Solve the following equation and check your answer:

$\dfrac{2x}{3} + 4x = 42$

Answer

We have:

$\phantom{=} \dfrac{2x}{3} + 4x = 42 \\[1em] \Rightarrow \dfrac{2x + 12x}{3} = 42 \\[1em] \Rightarrow 2x + 12x = 42 \times 3 \\[1em] \Rightarrow 14x = 126 \\[1em] \Rightarrow x = \dfrac{126}{14}$

∴ x = 9

Check:

LHS = $\dfrac{2x}{3} + 4x$

= $\dfrac{2(9)}{3} + 4(9)$

= $6 + 36$

= $42$

RHS = $42$

Hence, LHS = RHS.

Solve the following equation and check your answer:

x - 24% of x = 38

Answer

We have:

x - 24% of x = 38

$\Rightarrow x - \dfrac{24}{100}x = 38 \\[1em] \Rightarrow \dfrac{100x - 24x}{100} = 38 \\[1em] \Rightarrow \dfrac{76x}{100} = 38 \\[1em] \Rightarrow 76x = 38 \times 100 \\[1em] \Rightarrow 76x = 3800 \\[1em] \Rightarrow x = \dfrac{3800}{76}$

∴ x = 50

Check:

LHS = x - 24% of x

= 50 - 24% of 50

= 50 - 12

= 38

RHS = 38

Hence, LHS = RHS.

Solve the following equation and check your answer:

$\dfrac{x + 5}{2} + \dfrac{x}{3} = 20$

Answer

We have:

$\phantom{=} \dfrac{x + 5}{2} + \dfrac{x}{3} = 20 \\[1em] \Rightarrow \dfrac{3(x + 5) + 2x}{6} = 20 \\[1em] \Rightarrow 3x + 15 + 2x = 20 \times 6 \\[1em] \Rightarrow 5x + 15 = 120 \\[1em] \Rightarrow 5x = 120 - 15 \quad \text{[Transposing +15 to RHS]} \\[1em] \Rightarrow 5x = 105 \\[1em] \Rightarrow x = \dfrac{105}{5} \\[1em]$

∴ x = 21

Check:

LHS = $\dfrac{x + 5}{2} + \dfrac{x}{3}$

= $\dfrac{21 + 5}{2} + \dfrac{21}{3}$

= 13 + 7

= 20

RHS = 20

Hence, LHS = RHS.

Solve the following equation and check your answer:

$\dfrac{2x + 3}{3}-\dfrac{3x - 2}{4} = 1$

Answer

We have:

$\phantom{=} \dfrac{2x + 3}{3}-\dfrac{3x - 2}{4} = 1 \\[1em] \Rightarrow \dfrac{4(2x + 3) - 3(3x - 2)}{12} = 1 \\[1em] \Rightarrow 8x + 12 - 9x + 6 = 1 \times 12 \\[1em] \Rightarrow -x + 18 = 12 \\[1em] \Rightarrow -x = 12 - 18 \quad \text{[Transposing +18 to RHS]} \\[1em] \Rightarrow -x = -6 \\[1em] \Rightarrow x = 6 \\[1em]$

∴ x = 6

Check:

$\text{LHS} = \dfrac{2x + 3}{3} - \dfrac{3x - 2}{4} \\[1em] \phantom{\text{LHS}} = \dfrac{15}{3} - \dfrac{16}{4} \\[1em] \phantom{\text{LHS}} = 5 - 4 \\[1em] \phantom{\text{LHS}} = 1 \\[2em] \text{RHS} = 1$

Hence, LHS = RHS.

Solve the following equation and check your answer:

$\dfrac{3y - 2}{7} - \dfrac{5y - 8}{4} = \dfrac{1}{14}$

Answer

We have:

$\phantom{=} \dfrac{3y - 2}{7} - \dfrac{5y - 8}{4} = \dfrac{1}{14} \\[1em] \Rightarrow \dfrac{4(3y - 2) - 7(5y - 8)}{28} = \dfrac{1}{14} \\[1em] \Rightarrow 12y - 8 - 35y + 56 = \dfrac{1}{14} \times 28 \\[1em] \Rightarrow -23y + 48 = \dfrac{1}{1} \times 2 \\[1em] \Rightarrow -23y + 48 = 2 \\[1em] \Rightarrow -23y = 2 - 48 \quad \text{[Transposing +48 to RHS]} \\[1em] \Rightarrow -23y = - 46 \\[1em] y = \dfrac{-46}{-23} \\[1em]$

∴ y = 2

Check:

$\text{LHS} = \dfrac{3y - 2}{7} - \dfrac{5y - 8}{4} \\[1em] \phantom{\text{LHS}} = \dfrac{6 - 2}{7} - \dfrac{10 - 8}{4} \\[1em] \phantom{\text{LHS}} = \dfrac{4}{7} - \dfrac{1}{2} \\[1em] \phantom{\text{LHS}} = \dfrac{8 - 7}{14} \\[1em] \phantom{\text{LHS}} = \dfrac{1}{14} \\[2em] \text{RHS} = \dfrac{1}{14}$

Hence, LHS = RHS.

Solve the following equation and check your answer:

$\dfrac{x-2}{3}+\dfrac{x-3}{4} = \dfrac{x-1}{2}$

Answer

We have:

$\phantom{=} \dfrac{x-2}{3}+\dfrac{x-3}{4} = \dfrac{x-1}{2} \\[1em] \Rightarrow \dfrac{4(x - 2) + 3(x - 3)}{12} = \dfrac{x-1}{2} \\[1em] \Rightarrow 4x - 8 + 3x - 9 = \left(\dfrac{x-1}{2}\right) \times 12 \\[1em] \Rightarrow 7x - 17 = \left(\dfrac{x-1}{1}\right) \times 6 \\[1em] \Rightarrow 7x - 17 = 6(x-1) \\[1em] \Rightarrow 7x - 17 = 6x - 6 \\[1em] \Rightarrow 7x - 6x = -6 + 17 \\[1em] \quad \text{[Transposing -17 to RHS and +6x to LHS]}$

∴ x = 11

Check:

$\text{LHS} = \dfrac{x-2}{3} + \dfrac{x-3}{4} \\[1em] \phantom{\text{LHS}} = \dfrac{11-2}{3} + \dfrac{11-3}{4} \\[1em] \phantom{\text{LHS}} = 3 + 2 \\[1em] \phantom{\text{LHS}} = 5 \\[2em] \text{RHS} = \dfrac{x-1}{2} \\[1em] \phantom{\text{RHS}} = \dfrac{11-1}{2} \\[1em] \phantom{\text{RHS}} = 5$

Hence, LHS = RHS.

Solve the following equation and check your answer:

$\dfrac{x-4}{7}-\dfrac{x+4}{5} = \dfrac{x+3}{7}$

Answer

We have:

$\phantom{=} \dfrac{x-4}{7}-\dfrac{x+4}{5} = \dfrac{x+3}{7} \\[1em] \Rightarrow \dfrac{5(x - 4) - 7(x + 4)}{35} = \dfrac{x+3}{7} \\[1em] \Rightarrow 5x - 20 - 7x - 28 = \left(\dfrac{x+3}{7}\right) \times 35 \\[1em] \Rightarrow -2x - 48 = \left(\dfrac{x+3}{1}\right) \times 5 \\[1em] \Rightarrow -2x - 48 = 5(x+3) \\[1em] \Rightarrow -2x - 48 = 5x + 15 \\[1em] \Rightarrow -2x - 5x = 15 + 48 \\[1em] \quad \text{[Transposing -48 to RHS and +5x to LHS]} \\[1em] \Rightarrow -7x = 63 \\[1em] \Rightarrow -x = \dfrac{63}{7} \\[1em] \Rightarrow -x = 9$

∴ x = -9

Check:

$\text{LHS} = \dfrac{x-4}{7} - \dfrac{x+4}{5} \\[1em] \phantom{\text{LHS}} = \dfrac{-13}{7} - \dfrac{-5}{5} \\[1em] \phantom{\text{LHS}} = -\dfrac{13}{7} + 1 \\[1em] \phantom{\text{LHS}} = -\dfrac{6}{7} \\[2em] \text{RHS} = \dfrac{x+3}{7} \\[1em] \phantom{\text{RHS}} = \dfrac{-9+3}{7} \\[1em] \phantom{\text{RHS}} = -\dfrac{6}{7}$

Hence, LHS = RHS.

Solve the following equation and check your answer:

$\dfrac{2}{3}(3x - 2) = \dfrac{4}{5}(2x - 3)-\dfrac{4}{3}$

Answer

We have:

$\phantom{=} \dfrac{2}{3}(3x - 2) = \dfrac{4}{5}(2x - 3)-\dfrac{4}{3} \\[1em] \Rightarrow \dfrac{6x - 4}{3} = \dfrac{12(2x - 3) - 20}{15} \\[1em] \Rightarrow \dfrac{6x - 4}{3} = \dfrac{24x - 36 - 20}{15} \\[1em] \Rightarrow \dfrac{6x - 4}{3} = \dfrac{24x - 56}{15} \\[1em] \Rightarrow 15(6x - 4) = 3(24x - 56) \\[1em] \quad \text{[By cross multiplication]} \\[1em] \Rightarrow 90x - 60 = 72x - 168 \\[1em] \Rightarrow 90x - 72x = 60 - 168 \\[1em] \quad \text{[Transposing -60 to RHS and +72x to LHS]} \\[1em] \Rightarrow 18x = -108 \\[1em] \Rightarrow x = \dfrac{-108}{18} \\[1em]$

∴ x = -6

Check:

$\text{LHS} = \dfrac{2}{3}(3x - 2) \\[1em] \phantom{\text{LHS}} = \dfrac{2}{3}(-20) \\[1em] \phantom{\text{LHS}} = -\dfrac{40}{3} \\[2em] \text{RHS} = \dfrac{4}{5}(2x - 3) - \dfrac{4}{3} \\[1em] \phantom{\text{RHS}} = \dfrac{4}{5}(-15) - \dfrac{4}{3} \\[1em] \phantom{\text{RHS}} = -12 - \dfrac{4}{3} \\[1em] \phantom{\text{RHS}} = -\dfrac{40}{3}$

Hence, LHS = RHS.

Solve the following equation and check your answer:

$\dfrac{3}{4}(2x - 5) - \dfrac{5}{6}(7 - 5x) = \dfrac{7x}{3}$

Answer

We have:

$\phantom{=} \dfrac{3}{4}(2x - 5) - \dfrac{5}{6}(7 - 5x) = \dfrac{7x}{3} \\[1em] \Rightarrow \dfrac{9(2x - 5) - 10(7 - 5x)}{12} = \dfrac{7x}{3} \\[1em] \Rightarrow \dfrac{18x - 45 - 70 + 50x}{12} = \dfrac{7x}{3} \\[1em] \Rightarrow \dfrac{68x - 115}{12} = \dfrac{7x}{3} \\[1em] \Rightarrow 3(68x - 115) = 12(7x) \\[1em] \quad \text{[By cross multiplication]} \\[1em] \Rightarrow 68x - 115 = 4(7x) \\[1em] \quad \text{[Dividing by 3 on both sides]} \\[1em] \Rightarrow 68x - 115 = 28x \\[1em] \Rightarrow 68x - 28x = 115 \\[1em] \quad \text{[Transposing -115 to RHS and +28x to LHS]} \\[1em] \Rightarrow 40x = 115 \\[1em] \Rightarrow x = \dfrac{115}{40} \\[1em] \Rightarrow x = \dfrac{23}{8}$

∴ x = $2\dfrac{7}{8}$

Check:

$\text{LHS} = \dfrac{3}{4}(2x - 5) - \dfrac{5}{6}(7 - 5x) \\[1em] \phantom{\text{LHS}} = \dfrac{3}{4}\left(\dfrac{23}{4} - 5\right) - \dfrac{5}{6}\left(7 - \dfrac{115}{8}\right) \\[1em] \phantom{\text{LHS}} = \dfrac{3}{4}\left(\dfrac{3}{4}\right) - \dfrac{5}{6}\left(-\dfrac{59}{8}\right) \\[1em] \phantom{\text{LHS}} = \dfrac{9}{16} + \dfrac{295}{48} \\[1em] \phantom{\text{LHS}} = \dfrac{322}{48} \\[1em] \phantom{\text{LHS}} = \dfrac{161}{24} \\[2em] \text{RHS} = \dfrac{7x}{3} \\[1em] \phantom{\text{RHS}} = \dfrac{7}{3} \times \dfrac{23}{8} \\[1em] \phantom{\text{RHS}} = \dfrac{161}{24}$

Hence, LHS = RHS.

Solve the following equation and check your answer:

$x-\Big(2x - \dfrac{3x - 4}{7}\Big) = \dfrac{4x - 27}{3} - 3$

Answer

We have:

$\phantom{=} x-\Big(2x - \dfrac{3x - 4}{7}\Big) = \dfrac{4x - 27}{3} - 3 \\[1em] \Rightarrow x - 2x + \dfrac{3x - 4}{7} = \dfrac{4x - 27 - 9}{3} \\[1em] \Rightarrow -x + \dfrac{3x - 4}{7} = \dfrac{4x - 36}{3} \\[1em] \Rightarrow \dfrac{-7x + 3x - 4}{7} = \dfrac{4x - 36}{3} \\[1em] \Rightarrow \dfrac{-4x - 4}{7} = \dfrac{4x - 36}{3} \\[1em] \Rightarrow 3(-4x - 4) = 7(4x - 36) \\[1em] \quad \text{[By cross multiplication]} \\[1em] \Rightarrow -12x - 12 = 28x - 252 \\[1em] \Rightarrow -12x - 28x = 12 - 252 \\[1em] \quad \text{[Transposing -12 to RHS and +28x to LHS]} \\[1em] \Rightarrow -40x = -240 \\[1em] \Rightarrow x = \dfrac{-240}{-40}$

∴ x = 6

Check:

$\text{LHS} = x - \left(2x - \dfrac{3x - 4}{7}\right) \\[1em] \phantom{\text{LHS}} = 6 - \left(12 - 2\right) \\[1em] \phantom{\text{LHS}} = -4 \\[2em] \text{RHS} = \dfrac{4x - 27}{3} - 3 \\[1em] \phantom{\text{RHS}} = \dfrac{24 - 27}{3} - 3 \\[1em] \phantom{\text{RHS}} = -1 - 3 \\[1em] \phantom{\text{RHS}} = -4$

Hence, LHS = RHS.

Solve the following equation and check your answer:

$\dfrac{3}{4}(7x - 1)-\Big(2x -\dfrac{1-x}{2}\Big) = x + \dfrac{3}{2}$

Answer

We have:

$\phantom{=} \dfrac{3}{4}(7x - 1)-\Big(2x -\dfrac{1-x}{2}\Big) = x + \dfrac{3}{2} \\[1em] \Rightarrow \dfrac{21x - 3}{4}-\left(\dfrac{4x -(1 - x)}{2}\right) = \dfrac{2x + 3}{2} \\[1em] \Rightarrow \dfrac{21x - 3}{4}-\left(\dfrac{4x - 1 + x}{2}\right) = \dfrac{2x + 3}{2} \\[1em] \Rightarrow \dfrac{21x - 3}{4}-\dfrac{5x - 1}{2} = \dfrac{2x + 3}{2} \\[1em] \Rightarrow \dfrac{21x - 3 - 10x + 2}{4} = \dfrac{2x + 3}{2} \\[1em] \Rightarrow \dfrac{11x - 1}{4} = \dfrac{2x + 3}{2} \\[1em] \Rightarrow 2(11x - 1) = 4(2x + 3) \\[1em] \quad \text{[By cross multiplication]} \\[1em] \Rightarrow 22x - 2 = 8x + 12 \\[1em] \Rightarrow 22x - 8x = 12 + 2 \\[1em] \quad \text{[Transposing -2 to RHS and +8x to LHS]} \\[1em] \Rightarrow 14x = 14 \\[1em] \Rightarrow x = \dfrac{14}{14}$

∴ x = 1

Check:

$\text{LHS} = \dfrac{3}{4}(7x - 1) - \left(2x - \dfrac{1-x}{2}\right) \\[1em] \phantom{\text{LHS}} = \dfrac{3}{4}(6) - \left(2 - 0\right) \\[1em] \phantom{\text{LHS}} = \dfrac{9}{2} - 2 \\[1em] \phantom{\text{LHS}} = \dfrac{5}{2} \\[2em] \text{RHS} = x + \dfrac{3}{2} \\[1em] \phantom{\text{RHS}} = 1 + \dfrac{3}{2} \\[1em] \phantom{\text{RHS}} = \dfrac{5}{2}$

Hence, LHS = RHS.