Question

For each linear system of equations, identify the
solution; if there is no solution, put an asterisk (*) in
the boxes for the coordinate values.
System A has a solution at
System B has a solution at(
System C has a solution at
A. y=x+5
V = 3x -3
B.
-2V
-1+7
8V = 4x + 10
C. y=
-3x
x+ y = 16

Answer 1

EXPLANATION

First equation:

y = x + 5

y = 3x - 3

$\mathrm{Substitute\:}y=3x-3$$\begin{bmatrix}3x-3=x+5\end{bmatrix}$$\mathrm{Add\:}3\mathrm{\:to\:both\:sides}$$x-3+3=x+5+3$

Simplify:

$3x=x+8$$\mathrm{Subtract\:}x\mathrm{\:from\:both\:sides}$$3x-x=x+8-x$

Simplify:

$2x=8$$\mathrm{Divide\:both\:sides\:by\:}2$$\frac{2x}{2}=\frac{8}{2}$

Simplify:

$x=4$$\mathrm{For\:}y=3x-3$$\mathrm{Substitute\:}x=4$$y=3\cdot \:4-3$$\mathrm{Simplify}$$y=9$

Therefore, System A has a solution at (4,9)

Second equation:

$\begin{bmatrix}-2y=-x+7 \\ 8y=4x+10\end{bmatrix}$$\mathrm{Divide\:both\:sides\:by\:}-2$$\frac{-2y}{-2}=-\frac{x}{-2}+\frac{7}{-2}$

Simplifying:

$y=-\frac{-x+7}{2}$$\mathrm{Substitute\:}y=-\frac{-x+7}{2}$$\begin{bmatrix}8\left(-\frac{-x+7}{2}\right)=4x+10\end{bmatrix}$

Removing the parentheses:

$=-8\cdot \frac{-x+7}{2}$

Multiplying fractions:

$=-\frac{\left(-x+7\right)\cdot \:8}{2}$$\mathrm{Divide\:the\:numbers:}\:\frac{8}{2}=4$$-4\left(-x+7\right)=4x+10$$\mathrm{Refine\:}-4\left(-x+7\right)=4x+10$$-38=0$$-38=0\:\mathrm{\:is\:false,\:therefore\:the\:system\:of\:equations\:has\:no\:solution}$

System B has a solution at (*,*)

Third equation:

$\begin{bmatrix}y=-3x\\ x+y=16\end{bmatrix}$$\mathrm{Substitute\:}y=-3x$$\begin{bmatrix}x-3x=16\end{bmatrix}$$\begin{bmatrix}-2x=16\end{bmatrix}$$\mathrm{Divide\:both\:sides\:by\:}-2$$\frac{-2x}{-2}=\frac{16}{-2}$

Simplify the expression:

$x=-8$$\mathrm{For\:}y=-3x$$\mathrm{Substitute\:}x=-8$$y=-3\left(-8\right)$$\mathrm{Simplify}$$y=24$$\mathrm{The\:solutions\:to\:the\:system\:of\:equations\:are:}$$y=24,\:x=-8$

Therefore, System C has a solution at (-8,24)