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

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Solution

## (1) 17p − 2 = 49 $17p-2=49\phantom{\rule{0ex}{0ex}}\mathrm{Adding}2\mathrm{to}\mathrm{both}\mathrm{sides}\phantom{\rule{0ex}{0ex}}⇒17p-2+2=49+2\phantom{\rule{0ex}{0ex}}⇒17p=51\phantom{\rule{0ex}{0ex}}\mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}17\phantom{\rule{0ex}{0ex}}⇒\frac{17p}{17}=\frac{51}{17}\phantom{\rule{0ex}{0ex}}⇒p=3$ Hence, p = 3 is the solution of the equation. (2) 2m + 7 = 9 $2m+7=9\phantom{\rule{0ex}{0ex}}\mathrm{Subtracting}7\mathrm{from}\mathrm{both}\mathrm{sides}\phantom{\rule{0ex}{0ex}}⇒2m+7-7=9-7\phantom{\rule{0ex}{0ex}}⇒2m=2\phantom{\rule{0ex}{0ex}}\mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}2\phantom{\rule{0ex}{0ex}}⇒\frac{2m}{2}=\frac{2}{2}\phantom{\rule{0ex}{0ex}}⇒m=1$ Hence, m = 1 is the solution of the equation. (3) 3x + 12 = 2x − 4 $3x+12=2x-4\phantom{\rule{0ex}{0ex}}\mathrm{Subtracting}12\mathrm{from}\mathrm{both}\mathrm{sides}\phantom{\rule{0ex}{0ex}}⇒3x+12-12=2x-4-12\phantom{\rule{0ex}{0ex}}⇒3x=2x-16\phantom{\rule{0ex}{0ex}}\mathrm{Subtracting}2x\mathrm{from}\mathrm{both}\mathrm{sides}\phantom{\rule{0ex}{0ex}}⇒3x-2x=2x-16-2x\phantom{\rule{0ex}{0ex}}⇒x=-16$ Hence, x = −16 is the solution of the equation. (4) 5(x − 3) = 3(x + 2) $5\left(x-3\right)=3\left(x+2\right)\phantom{\rule{0ex}{0ex}}⇒5x-15=3x+6\phantom{\rule{0ex}{0ex}}\mathrm{Adding}15\mathrm{to}\mathrm{both}\mathrm{sides}\phantom{\rule{0ex}{0ex}}⇒5x-15+15=3x+6+15\phantom{\rule{0ex}{0ex}}⇒5x=3x+21\phantom{\rule{0ex}{0ex}}\mathrm{Subtracting}3\mathrm{x}\mathrm{from}\mathrm{both}\mathrm{sides}\phantom{\rule{0ex}{0ex}}⇒5x-3x=3x+21-3x\phantom{\rule{0ex}{0ex}}⇒2x=21\phantom{\rule{0ex}{0ex}}\mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}2\phantom{\rule{0ex}{0ex}}⇒\frac{2x}{2}=\frac{21}{2}\phantom{\rule{0ex}{0ex}}⇒x=\frac{21}{2}$ Hence, x = $\frac{21}{2}$ is the solution of the equation. (5) $\frac{9x}{8}$+ 1 = 10 $\frac{9x}{8}+1=10\phantom{\rule{0ex}{0ex}}\mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{by}8\phantom{\rule{0ex}{0ex}}⇒\frac{9x}{8}×8+1×8=10×8\phantom{\rule{0ex}{0ex}}⇒9x+8=80\phantom{\rule{0ex}{0ex}}\mathrm{Subtracting}8\mathrm{from}\mathrm{both}\mathrm{sides}\phantom{\rule{0ex}{0ex}}⇒9x+8-8=80-8\phantom{\rule{0ex}{0ex}}⇒9x=72\phantom{\rule{0ex}{0ex}}\mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}9\phantom{\rule{0ex}{0ex}}⇒\frac{9x}{9}=\frac{72}{9}\phantom{\rule{0ex}{0ex}}⇒x=8$ Hence, x = 8 is the solution of the equation. (6) $\frac{y}{7}$ + $\frac{y-4}{3}$ = 2 $\frac{y}{7}+\frac{y-4}{3}=2\phantom{\rule{0ex}{0ex}}\mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{by}21\phantom{\rule{0ex}{0ex}}⇒21\left(\frac{y}{7}\right)+21\left(\frac{y-4}{3}\right)=2×21\phantom{\rule{0ex}{0ex}}⇒3\left(y\right)+7\left(y-4\right)=42\phantom{\rule{0ex}{0ex}}⇒3y+7y-28=42\phantom{\rule{0ex}{0ex}}⇒10y-28=42\phantom{\rule{0ex}{0ex}}\mathrm{Adding}28\mathrm{to}\mathrm{both}\mathrm{sides}\phantom{\rule{0ex}{0ex}}⇒10y-28+28=42+28\phantom{\rule{0ex}{0ex}}⇒10y=70\phantom{\rule{0ex}{0ex}}\mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}10\phantom{\rule{0ex}{0ex}}⇒\frac{10y}{10}=\frac{70}{10}\phantom{\rule{0ex}{0ex}}⇒y=7$ Hence, y = 7 is the solution of the equation. (7) 13x − 5 = $\frac{3}{2}$ $13x-5=\frac{3}{2}\phantom{\rule{0ex}{0ex}}\mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{by}2\phantom{\rule{0ex}{0ex}}⇒2\left(13x-5\right)=2×\frac{3}{2}\phantom{\rule{0ex}{0ex}}⇒26x-10=3\phantom{\rule{0ex}{0ex}}\mathrm{Adding}10\mathrm{to}\mathrm{both}\mathrm{sides}\phantom{\rule{0ex}{0ex}}⇒26x-10+10=3+10\phantom{\rule{0ex}{0ex}}⇒26x=13\phantom{\rule{0ex}{0ex}}\mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}26\phantom{\rule{0ex}{0ex}}⇒\frac{26x}{26}=\frac{13}{26}\phantom{\rule{0ex}{0ex}}⇒x=\frac{1}{2}$ Hence, x = $\frac{1}{2}$ is the solution of the equation. (8) 3(y + 8) = 10(y − 4) + 8 $3\left(y+8\right)=10\left(y-4\right)+8\phantom{\rule{0ex}{0ex}}⇒3y+24=10y-40+8\phantom{\rule{0ex}{0ex}}⇒3y+24=10y-32\phantom{\rule{0ex}{0ex}}\mathrm{Adding}32\mathrm{to}\mathrm{both}\mathrm{sides}\phantom{\rule{0ex}{0ex}}⇒3y+24+32=10y-32+32\phantom{\rule{0ex}{0ex}}⇒3y+56=10y\phantom{\rule{0ex}{0ex}}\mathrm{Subtracting}3y\mathrm{from}\mathrm{both}\mathrm{sides}\phantom{\rule{0ex}{0ex}}⇒3y+56-3y=10y-3y\phantom{\rule{0ex}{0ex}}⇒56=7y\phantom{\rule{0ex}{0ex}}\mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}7\phantom{\rule{0ex}{0ex}}⇒\frac{7y}{7}=\frac{56}{7}\phantom{\rule{0ex}{0ex}}⇒y=8$ Hence, y = 8 is the solution of the equation. (9) $\frac{x-9}{x-5}$ = $\frac{5}{7}$ $\frac{x-9}{x-5}=\frac{5}{7}\phantom{\rule{0ex}{0ex}}\mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{by}7\phantom{\rule{0ex}{0ex}}⇒7\left(\frac{x-9}{x-5}\right)=\frac{5}{7}×7\phantom{\rule{0ex}{0ex}}⇒\frac{7x-63}{x-5}=5\phantom{\rule{0ex}{0ex}}\mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{by}\left(x-5\right)\phantom{\rule{0ex}{0ex}}⇒\frac{7x-63}{x-5}\left(x-5\right)=5\left(x-5\right)\phantom{\rule{0ex}{0ex}}⇒7x-63=5x-25\phantom{\rule{0ex}{0ex}}\mathrm{Adding}63\mathrm{to}\mathrm{both}\mathrm{sides}\phantom{\rule{0ex}{0ex}}⇒7x-63+63=5x-25+63\phantom{\rule{0ex}{0ex}}⇒7x=5x+38\phantom{\rule{0ex}{0ex}}\mathrm{Subtracting}5x\mathrm{from}\mathrm{both}\mathrm{sides}\phantom{\rule{0ex}{0ex}}⇒7x-5x=5x+38-5x\phantom{\rule{0ex}{0ex}}⇒2x=38\phantom{\rule{0ex}{0ex}}\mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}2\phantom{\rule{0ex}{0ex}}⇒\frac{2x}{2}=\frac{38}{2}\phantom{\rule{0ex}{0ex}}⇒x=19$ Hence, x = 19 is the solution of the equation. (10) $\frac{y-4}{3}$ + 3y = 4 $\frac{y-4}{3}+3y=4\phantom{\rule{0ex}{0ex}}\mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{by}3\phantom{\rule{0ex}{0ex}}⇒3\left(\frac{y-4}{3}\right)+3\left(3y\right)=4×3\phantom{\rule{0ex}{0ex}}⇒y-4+9y=12\phantom{\rule{0ex}{0ex}}⇒10y-4=12\phantom{\rule{0ex}{0ex}}\mathrm{Adding}4\mathrm{to}\mathrm{both}\mathrm{sides}\phantom{\rule{0ex}{0ex}}⇒10y-4+4=12+4\phantom{\rule{0ex}{0ex}}⇒10y=16\phantom{\rule{0ex}{0ex}}\mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}10\phantom{\rule{0ex}{0ex}}⇒\frac{10y}{10}=\frac{16}{10}\phantom{\rule{0ex}{0ex}}⇒y=\frac{8}{5}$ Hence, y = $\frac{8}{5}$ is the solution of the equation. (11) $\frac{b+\left(b+1\right)+\left(b+2\right)}{4}$ = 21 $\frac{b+\left(b+1\right)+\left(b+2\right)}{4}=21\phantom{\rule{0ex}{0ex}}\mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{by}4\phantom{\rule{0ex}{0ex}}⇒4×\frac{b+\left(b+1\right)+\left(b+2\right)}{4}=4×21\phantom{\rule{0ex}{0ex}}⇒b+\left(b+1\right)+\left(b+2\right)=84\phantom{\rule{0ex}{0ex}}⇒3b+3=84\phantom{\rule{0ex}{0ex}}\mathrm{Subtracting}3\mathrm{from}\mathrm{both}\mathrm{sides}\phantom{\rule{0ex}{0ex}}⇒3b+3-3=84-3\phantom{\rule{0ex}{0ex}}⇒3b=81\phantom{\rule{0ex}{0ex}}\mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}3\phantom{\rule{0ex}{0ex}}⇒\frac{3b}{3}=\frac{81}{3}\phantom{\rule{0ex}{0ex}}⇒b=27$ Hence, b = 27 is the solution of the equation.

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