Question

3x/2-5y/3=-2,x/2+y/2=13/6 solve by substitution method

Answer 1

Answer:

$3\cdot \dfrac{x}{2}-5\cdot \dfrac{y}{3}=-2,\:\dfrac{x}{2}+\dfrac{y}{2}=\dfrac{13}{6}\quad :\quad y=\dfrac{51}{19},\:x=\dfrac{94}{57}$

Step-by-step explanation:

$\begin{bmatrix}3\cdot \dfrac{x}{2}-5\cdot \dfrac{y}{3}=-2\\ \dfrac{x}{2}+\dfrac{y}{2}=\dfrac{13}{6}\end{bmatrix}$

$\black{\mathrm{Isolate}\:x\:\mathrm{for}\:3\dfrac{x}{2}-5\dfrac{y}{3}=-2:}$

$3\cdot \dfrac{x}{2}-5\cdot \dfrac{y}{3}=-2$

$\gray{\mathrm{Add\:}5\dfrac{y}{3}\mathrm{\:to\:both\:sides}}$

$3\cdot \dfrac{x}{2}-5\cdot \dfrac{y}{3}+5\cdot \dfrac{y}{3}=-2+5\cdot \dfrac{y}{3}$

$\gray{\mathrm{Simplify}}$

$3\cdot \dfrac{x}{2}=-2+\dfrac{5y}{3}$

$\gray{\mathrm{Refine\:}3\cdot \dfrac{x}{2}:\quad \dfrac{3x}{2}}$

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

$\gray{\mathrm{Multiply\:both\:sides\:by\:}2}$

$\dfrac{3x}{2}\cdot \:2=-2\cdot \:2+\dfrac{5y}{3}\cdot \:2$

$\gray{\mathrm{Simplify}}$

$3x=-4+\dfrac{10y}{3}$

$\gray{\mathrm{Divide\:both\:sides\:by\:}3}$

$\dfrac{3x}{3}=-\dfrac{4}{3}+\dfrac{\dfrac{10y}{3}}{3}$

$\gray{\mathrm{Simplify}}$

$x=\dfrac{-12+10y}{9}$

$\gray{\mathrm{Subsititute\:}x=\dfrac{-12+10y}{9}}$

$\begin{bmatrix}\dfrac{\dfrac{-12+10y}{9}}{2}+\dfrac{y}{2}=\dfrac{13}{6}\end{bmatrix}$

$\black{\mathrm{Isolate}\:y\:\mathrm{for}\:\dfrac{\dfrac{-12+10y}{9}}{2}+\dfrac{y}{2}=\dfrac{13}{6}:}$

$\dfrac{\dfrac{-12+10y}{9}}{2}+\dfrac{y}{2}=\dfrac{13}{6}$

$\gray{\mathrm{Expand\:}\dfrac{\dfrac{-12+10y}{9}}{2}+\dfrac{y}{2}:\quad -\dfrac{2}{3}+\dfrac{19y}{18}}$

$-\dfrac{2}{3}+\dfrac{19y}{18}=\dfrac{13}{6}$

$\gray{\mathrm{Multiply\:both\:sides\:by\:}18}$

$-\dfrac{2}{3}\cdot \:18+\dfrac{19y}{18}\cdot \:18=\dfrac{13}{6}\cdot \:18$

$\gray{\mathrm{Simplify}}$

$-12+19y=39$

$\gray{\mathrm{Add\:}12\mathrm{\:to\:both\:sides}}$

$-12+19y+12=39+12$

$\gray{\mathrm{Simplify}}$

$19y=51$

$\gray{\mathrm{Divide\:both\:sides\:by\:}19}$

$\dfrac{19y}{19}=\dfrac{51}{19}$

$\gray{\mathrm{Simplify}}$

$y=\dfrac{51}{19}$

$\gray{\mathrm{For\:}x=\dfrac{-12+10y}{9}}$

$\gray{\mathrm{Subsititute\:}y=\dfrac{51}{19}}$

$\gray{x=\dfrac{-12+10\cdot \dfrac{51}{19}}{9}}$

$\black{\dfrac{-12+10\cdot \dfrac{51}{19}}{9}}$

$\gray{10\cdot \dfrac{51}{19}=\dfrac{510}{19}}$

$=\dfrac{-12+\dfrac{510}{19}}{9}$

$\gray{\mathrm{Join}\:-12+\dfrac{510}{19}:\quad \dfrac{282}{19}}$

$=\dfrac{\dfrac{282}{19}}{9}$

$\gray{\mathrm{Apply\:the\:fraction\:rule}:\quad \dfrac{\dfrac{b}{c}}{a}=\dfrac{b}{c\:\cdot \:a}}$

$=\dfrac{282}{19\cdot \:9}$

$\gray{\mathrm{Multiply\:the\:numbers:}\:19\cdot \:9=171}$

$=\dfrac{282}{171}$

$\gray{\mathrm{Cancel\:the\:common\:factor:}\:3}$

$=\dfrac{94}{57}$

$x=\dfrac{94}{57}$

$\gray{\mathrm{The\:solutions\:to\:the\:system\:of\:equations\:are:}}$

$y=\dfrac{51}{19},\:x=\dfrac{94}{57}$