Question

Solve the following equations.
(1) 17p − 2 = 49
(2) 2m + 7 = 9
(3) 3x + 12 = 2x − 4
(4) 5(x − 3) = 3(x + 2)

(5) $\frac{9x}{8}$+ 1 = 10

(6) $\frac{y}{7}$ + $\frac{y-4}{3}$ = 2

(7) 13x − 5 = $\frac{3}{2}$

(8) 3(y + 8) = 10(y − 4) + 8

(9) $\frac{x-9}{x-5}$ = $\frac{5}{7}$

(10) $\frac{y-4}{3}$ + 3y = 4

(11) $\frac{b+(b+1)+(b+2)}{4}$ = 21

Answer 1

(1) 17p − 2 = 49

$$\begin{gathered} 17p-2=49 \\ \mathrm{Adding}2\mathrm{to}\mathrm{both}\mathrm{sides} \\ \Rightarrow 17p-2+2=49+2 \\ \Rightarrow 17p=51 \\ \mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}17 \\ \Rightarrow \frac{17p}{17}=\frac{51}{17} \\ \Rightarrow p=3 \end{gathered}$$

Hence, p = 3 is the solution of the equation.

(2) 2m + 7 = 9

$$\begin{gathered} 2m+7=9 \\ \mathrm{Subtracting}7\mathrm{from}\mathrm{both}\mathrm{sides} \\ \Rightarrow 2m+7-7=9-7 \\ \Rightarrow 2m=2 \\ \mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}2 \\ \Rightarrow \frac{2m}{2}=\frac{2}{2} \\ \Rightarrow m=1 \end{gathered}$$

Hence, m = 1 is the solution of the equation.

(3) 3x + 12 = 2x − 4

$$\begin{gathered} 3x+12=2x-4 \\ \mathrm{Subtracting}12\mathrm{from}\mathrm{both}\mathrm{sides} \\ \Rightarrow 3x+12-12=2x-4-12 \\ \Rightarrow 3x=2x-16 \\ \mathrm{Subtracting}2x\mathrm{from}\mathrm{both}\mathrm{sides} \\ \Rightarrow 3x-2x=2x-16-2x \\ \Rightarrow x=-16 \end{gathered}$$

Hence, x = −16 is the solution of the equation.

(4) 5(x − 3) = 3(x + 2)

$$\begin{gathered} 5\left(x-3\right)=3\left(x+2\right) \\ \Rightarrow 5x-15=3x+6 \\ \mathrm{Adding}15\mathrm{to}\mathrm{both}\mathrm{sides} \\ \Rightarrow 5x-15+15=3x+6+15 \\ \Rightarrow 5x=3x+21 \\ \mathrm{Subtracting}3\mathrm{x}\mathrm{from}\mathrm{both}\mathrm{sides} \\ \Rightarrow 5x-3x=3x+21-3x \\ \Rightarrow 2x=21 \\ \mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}2 \\ \Rightarrow \frac{2x}{2}=\frac{21}{2} \\ \Rightarrow x=\frac{21}{2} \end{gathered}$$

Hence, x = $\frac{21}{2}$ is the solution of the equation.

(5) $\frac{9x}{8}$+ 1 = 10

$$\begin{gathered} \frac{9x}{8}+1=10 \\ \mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{by}8 \\ \Rightarrow \frac{9x}{8}\times 8+1\times 8=10\times 8 \\ \Rightarrow 9x+8=80 \\ \mathrm{Subtracting}8\mathrm{from}\mathrm{both}\mathrm{sides} \\ \Rightarrow 9x+8-8=80-8 \\ \Rightarrow 9x=72 \\ \mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}9 \\ \Rightarrow \frac{9x}{9}=\frac{72}{9} \\ \Rightarrow x=8 \end{gathered}$$

Hence, x = 8 is the solution of the equation.

(6) $\frac{y}{7}$ + $\frac{y-4}{3}$ = 2

$$\begin{gathered} \frac{y}{7}+\frac{y-4}{3}=2 \\ \mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{by}21 \\ \Rightarrow 21\left(\frac{y}{7}\right)+21\left(\frac{y-4}{3}\right)=2\times 21 \\ \Rightarrow 3\left(y\right)+7\left(y-4\right)=42 \\ \Rightarrow 3y+7y-28=42 \\ \Rightarrow 10y-28=42 \\ \mathrm{Adding}28\mathrm{to}\mathrm{both}\mathrm{sides} \\ \Rightarrow 10y-28+28=42+28 \\ \Rightarrow 10y=70 \\ \mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}10 \\ \Rightarrow \frac{10y}{10}=\frac{70}{10} \\ \Rightarrow y=7 \end{gathered}$$

Hence, y = 7 is the solution of the equation.

(7) 13x − 5 = $\frac{3}{2}$

$$\begin{gathered} 13x-5=\frac{3}{2} \\ \mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{by}2 \\ \Rightarrow 2\left(13x-5\right)=2\times \frac{3}{2} \\ \Rightarrow 26x-10=3 \\ \mathrm{Adding}10\mathrm{to}\mathrm{both}\mathrm{sides} \\ \Rightarrow 26x-10+10=3+10 \\ \Rightarrow 26x=13 \\ \mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}26 \\ \Rightarrow \frac{26x}{26}=\frac{13}{26} \\ \Rightarrow x=\frac{1}{2} \end{gathered}$$

Hence, x = $\frac{1}{2}$ is the solution of the equation.

(8) 3(y + 8) = 10(y − 4) + 8

$$\begin{gathered} 3\left(y+8\right)=10\left(y-4\right)+8 \\ \Rightarrow 3y+24=10y-40+8 \\ \Rightarrow 3y+24=10y-32 \\ \mathrm{Adding}32\mathrm{to}\mathrm{both}\mathrm{sides} \\ \Rightarrow 3y+24+32=10y-32+32 \\ \Rightarrow 3y+56=10y \\ \mathrm{Subtracting}3y\mathrm{from}\mathrm{both}\mathrm{sides} \\ \Rightarrow 3y+56-3y=10y-3y \\ \Rightarrow 56=7y \\ \mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}7 \\ \Rightarrow \frac{7y}{7}=\frac{56}{7} \\ \Rightarrow y=8 \end{gathered}$$

Hence, y = 8 is the solution of the equation.

(9) $\frac{x-9}{x-5}$ = $\frac{5}{7}$

$$\begin{gathered} \frac{x-9}{x-5}=\frac{5}{7} \\ \mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{by}7 \\ \Rightarrow 7\left(\frac{x-9}{x-5}\right)=\frac{5}{7}\times 7 \\ \Rightarrow \frac{7x-63}{x-5}=5 \\ \mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{by}\left(x-5\right) \\ \Rightarrow \frac{7x-63}{x-5}\left(x-5\right)=5\left(x-5\right) \\ \Rightarrow 7x-63=5x-25 \\ \mathrm{Adding}63\mathrm{to}\mathrm{both}\mathrm{sides} \\ \Rightarrow 7x-63+63=5x-25+63 \\ \Rightarrow 7x=5x+38 \\ \mathrm{Subtracting}5x\mathrm{from}\mathrm{both}\mathrm{sides} \\ \Rightarrow 7x-5x=5x+38-5x \\ \Rightarrow 2x=38 \\ \mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}2 \\ \Rightarrow \frac{2x}{2}=\frac{38}{2} \\ \Rightarrow x=19 \end{gathered}$$

Hence, x = 19 is the solution of the equation.

(10) $\frac{y-4}{3}$ + 3y = 4

$$\begin{gathered} \frac{y-4}{3}+3y=4 \\ \mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{by}3 \\ \Rightarrow 3\left(\frac{y-4}{3}\right)+3\left(3y\right)=4\times 3 \\ \Rightarrow y-4+9y=12 \\ \Rightarrow 10y-4=12 \\ \mathrm{Adding}4\mathrm{to}\mathrm{both}\mathrm{sides} \\ \Rightarrow 10y-4+4=12+4 \\ \Rightarrow 10y=16 \\ \mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}10 \\ \Rightarrow \frac{10y}{10}=\frac{16}{10} \\ \Rightarrow y=\frac{8}{5} \end{gathered}$$

Hence, y = $\frac{8}{5}$ is the solution of the equation.

(11) $\frac{b+(b+1)+(b+2)}{4}$ = 21

$$\begin{gathered} \frac{b+\left(b+1\right)+\left(b+2\right)}{4}=21 \\ \mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{by}4 \\ \Rightarrow 4\times \frac{b+\left(b+1\right)+\left(b+2\right)}{4}=4\times 21 \\ \Rightarrow b+\left(b+1\right)+\left(b+2\right)=84 \\ \Rightarrow 3b+3=84 \\ \mathrm{Subtracting}3\mathrm{from}\mathrm{both}\mathrm{sides} \\ \Rightarrow 3b+3-3=84-3 \\ \Rightarrow 3b=81 \\ \mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}3 \\ \Rightarrow \frac{3b}{3}=\frac{81}{3} \\ \Rightarrow b=27 \end{gathered}$$

Hence, b = 27 is the solution of the equation.