Question

solve the equation 3x+y+4=0 , 6x-2y+4=0 by substitution

Answer 1

Step-by-step explanation:

check this image.

here is your answer.

Answer 2

Answer:

$3x+y+4=0,\:6x-2y+4=0\quad :\quad y=-1,\:x=-1$

Step-by-step explanation:

$\begin{bmatrix}3x+y+4=0\\ 6x-2y+4=0\end{bmatrix}$

$\mathrm{Isolate}\:x\:\mathrm{for}\:3x+y+4=0$

$3x+y+4=0$

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

$3x+y+4-y=0-y$

$\mathrm{Simplify}$

$3x+4=-y$

$\mathrm{Subtract\:}4\mathrm{\:from\:both\:sides}$

$3x+4-4=-y-4$

$\mathrm{Simplify}$

$3x=-y-4$

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

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

$\mathrm{Simplify}$

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

$\mathrm{Subsititute\:}x=\dfrac{-y-4}{3}$

$\begin{bmatrix}6\cdot \dfrac{-y-4}{3}-2y+4=0\end{bmatrix}$

$\mathrm{Isolate}\:y\:\mathrm{for}\:6\dfrac{-y-4}{3}-2y+4=0$

$6\cdot \dfrac{-y-4}{3}-2y+4=0$

$6\cdot \dfrac{-y-4}{3}=2\left(-y-4\right)$

$2\left(-y-4\right)-2y+4=0$

$\mathrm{Subtract\:}4\mathrm{\:from\:both\:sides}$

$2\left(-y-4\right)-2y+4-4=0-4$

$\mathrm{Simplify}$

$2\left(-y-4\right)-2y=-4$

$\mathrm{Expand\:}2\left(-y-4\right)-2y$

$2\left(-y-4\right)-2y$

$\mathrm{Expand}\:2\left(-y-4\right):\quad -2y-8$

$=-2y-8-2y$

$\mathrm{Simplify}\:-2y-8-2y$

$\mathrm{Group\:like\:terms}$

$=-2y-2y-8$

$\mathrm{Add\:similar\:elements:}\:-2y-2y=-4y$

$=-4y-8$

$-4y-8=-4$

$\mathrm{Add\:}8\mathrm{\:to\:both\:sides}$

$-4y-8+8=-4+8$

$\mathrm{Simplify}$

$-4y=4$

$\mathrm{Divide\:both\:sides\:by\:}-4$

$\dfrac{-4y}{-4}=\dfrac{4}{-4}$

$\mathrm{Simplify}$

$y=-1$

$\mathrm{For\:}x=\dfrac{-y-4}{3}$

$\mathrm{Subsititute\:}y=-1$

$x=\dfrac{-\left(-1\right)-4}{3}$

$\dfrac{-\left(-1\right)-4}{3}$

$\mathrm{Apply\:rule}\:-\left(-a\right)=a$

$=\dfrac{1-4}{3}$

$\mathrm{Subtract\:the\:numbers:}\:1-4=-3$

$=\dfrac{-3}{3}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \dfrac{-a}{b}=-\dfrac{a}{b}$

$=-\dfrac{3}{3}$

$\mathrm{Apply\:rule}\:\dfrac{a}{a}=1$

$=-1$

$x=-1$

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

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