Question

system of equations graphically:
3x + y + 1 = 0
2x - 3y + 8 = 0​

Answer 1

Answer:

$3x+y+1=0,\:2x-3y+8=0\quad :\quad y=2,\:x=-1$

Step-by-step explanation:

$\begin{bmatrix}3x+y+1=0\\ 2x-3y+8=0\end{bmatrix}$

$\mathrm{Isolate}\:x\:\mathrm{for}\:3x+y+1=0$

$3x+y+1=0$

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

$3x+y+1-y=0-y$

$\gray{\mathrm{Simplify}}$

$3x+1=-y$

$\gray{\mathrm{Subtract\:}1\mathrm{\:from\:both\:sides}}$

$3x+1-1=-y-1$

$\gray{\mathrm{Simplify}}$

$3x=-y-1$

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

$\dfrac{3x}{3}=-\dfrac{y}{3}-\dfrac{1}{3}$

$\gray{\mathrm{Simplify}}$

$x=\dfrac{-y-1}{3}$

$\gray{\mathrm{Subsititute\:}x=\dfrac{-y-1}{3}}$

$\begin{bmatrix}2\cdot \dfrac{-y-1}{3}-3y+8=0\end{bmatrix}$

$\mathrm{Isolate}\:y\:\mathrm{for}\:2\cdot\dfrac{-y-1}{3}-3y+8=0$

$2\cdot \dfrac{-y-1}{3}-3y+8=0$

$\gray{\mathrm{Subtract\:}8\mathrm{\:from\:both\:sides}}$

$2\cdot \dfrac{-y-1}{3}-3y+8-8=0-8$

$\gray{\mathrm{Simplify}}$

$2\cdot \dfrac{-y-1}{3}-3y=-8$

$\mathrm{Expand\:}2\cdot \dfrac{-y-1}{3}-3y:\quad -\dfrac{11y}{3}-\dfrac{2}{3}$

$-\dfrac{11y}{3}-\dfrac{2}{3}=-8$

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

$-\dfrac{11y}{3}\cdot \:3-\dfrac{2}{3}\cdot \:3=-8\cdot \:3$

$\gray{\mathrm{Simplify}}$

$-11y-2=-24$

$\gray{\mathrm{Divide\:both\:sides\:by\:}-11}$

$\dfrac{-11y}{-11}=\dfrac{-22}{-11}$

$\gray{\mathrm{Simplify}}$

$y=2$

$\gray{\mathrm{For\:}x=\dfrac{-y-1}{3}}$

$\gray{\mathrm{Subsititute\:}y=2}$

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

$\gray{\dfrac{-2-1}{3}=-1}$

$x=-1$

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

$y=2,\:x=-1$