Question

1. What is the solution of the equations 2x - 3y = 7 and
4x - y = 20?
(1)​

Answer 1

Answer:

$2x-3y=7,\:4x-y=20\quad :\quad y=\dfrac{6}{5},\:x=\dfrac{53}{10}$

Step-by-step explanation:

$\begin{bmatrix}2x-3y=7\\ 4x-y=20\end{bmatrix}$

$\mathrm{Isolate}\:x\:\mathrm{for}\:2x-3y=7$

$2x-3y=7$

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

$2x-3y+3y=7+3y$

$\gray{\mathrm{Simplify}}$

$2x=7+3y$

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

$\dfrac{2x}{2}=\dfrac{7}{2}+\dfrac{3y}{2}$

$\gray{\mathrm{Simplify}}$

$x=\dfrac{7+3y}{2}$

$\gray{\mathrm{Subsititute\:}x=\dfrac{7+3y}{2}}$

$\begin{bmatrix}4\cdot \dfrac{7+3y}{2}-y=20\end{bmatrix}$

$\mathrm{Isolate}\:y\:\mathrm{for}\:4\cdot\dfrac{7+3y}{2}-y=20$

$4\cdot \dfrac{7+3y}{2}-y=20$

$\gray{4\cdot \dfrac{7+3y}{2}=2\left(3y+7\right)}$

$2\left(3y+7\right)-y=20$

$\gray{\mathrm{Expand\:}2\left(3y+7\right)-y:\quad 5y+14}$

$5y+14=20$

$\gray{\mathrm{Subtract\:}14\mathrm{\:from\:both\:sides}}$

$5y+14-14=20-14$

$\gray{\mathrm{Simplify}}$

$5y=6$

$\gray{\mathrm{Divide\:both\:sides\:by\:}5}$

$\dfrac{5y}{5}=\dfrac{6}{5}$

$\gray{\mathrm{Simplify}}$

$y=\dfrac{6}{5}$

$\gray{\mathrm{For\:}x=\dfrac{7+3y}{2}}$

$\gray{\mathrm{Subsititute\:}y=\dfrac{6}{5}}$

$x=\dfrac{7+3\cdot \dfrac{6}{5}}{2}$

$\gray{\dfrac{7+3\cdot \dfrac{6}{5}}{2}}$

$=\dfrac{7+\dfrac{18}{5}}{2}$

$=\dfrac{\dfrac{53}{5}}{2}$

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

$=\dfrac{53}{5\cdot \:2}$

$\gray{\mathrm{Multiply\:the\:numbers:}\:5\cdot \:2=10}$

$=\dfrac{53}{10}$

$x=\dfrac{53}{10}$

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

$y=\dfrac{6}{5},\:x=\dfrac{53}{10}$