Question

3x +2y + 25 = 0,
2x + y + 10 = 0.​

Answer 1

Answer:

$\green{3x+2y+25=0,\:2x+y+10=0\quad :\quad y=-20,\:x=5}$

Step-by-step explanation:

$\begin{bmatrix}3x+2y+25=0\\ 2x+y+10=0\end{bmatrix}$

$\black{\mathrm{Isolate}\:x\:\mathrm{for}\:3x+2y+25=0:}$

$3x+2y+25=0$

$\gray{\mathrm{Subtract\:}2y\mathrm{\:from\:both\:sides}}$

$3x+2y+25-2y=0-2y$

$\gray{\mathrm{Simplify}}$

$3x+25=-2y$

$\gray{\mathrm{Subtract\:}25\mathrm{\:from\:both\:sides}}$

$3x+25-25=-2y-25$

$\gray{\mathrm{Simplify}}$

$3x=-2y-25$

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

$\frac{3x}{3}=-\frac{2y}{3}-\frac{25}{3}$

$\gray{\mathrm{Simplify}}$

$x=\frac{-2y-25}{3}$

$\gray{\mathrm{Subsititute\:}x=\frac{-2y-25}{3}}$

$\begin{bmatrix}2\cdot \frac{-2y-25}{3}+y+10=0\end{bmatrix}$

$\black{\mathrm{Isolate}\:y\:\mathrm{for}\:2\frac{-2y-25}{3}+y+10=0:}$

$2\cdot \frac{-2y-25}{3}+y+10=0$

$\gray{\mathrm{Subtract\:}10\mathrm{\:from\:both\:sides}}$

$2\cdot \frac{-2y-25}{3}+y+10-10=0-10$

$\gray{\mathrm{Simplify}}$

$2\cdot \frac{-2y-25}{3}+y=-10$

$\gray{\mathrm{Expand\:}2\cdot \frac{-2y-25}{3}+y:\quad -\frac{y}{3}-\frac{50}{3}}$

$-\frac{y}{3}-\frac{50}{3}=-10$

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

$-\frac{y}{3}\cdot \:3-\frac{50}{3}\cdot \:3=-10\cdot \:3$

$\gray{\mathrm{Simplify}}$

$-y-50=-30$

$\gray{\mathrm{Add\:}50\mathrm{\:to\:both\:sides}}$

$-y-50+50=-30+50$

$\gray{\mathrm{Simplify}}$

$-y=20$

$\gray{\mathrm{Divide\:both\:sides\:by\:}-1}$

$\frac{-y}{-1}=\frac{20}{-1}$

$\gray{\mathrm{Simplify}}$

$y=-20$

$\gray{\mathrm{For\:}x=\frac{-2y-25}{3}}$

$\gray{\mathrm{Subsititute\:}y=-20}$

$x=\frac{-2\left(-20\right)-25}{3}$

$\black{\frac{-2\left(-20\right)-25}{3}}$

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

$=\frac{2\cdot \:20-25}{3}$

$\gray{2\cdot \:20-25=15}$

$=\frac{15}{3}$

$\gray{\mathrm{Divide\:the\:numbers:}\:\frac{15}{3}=5}$

$=5$

$x=5$

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

$y=-20,\:x=5$