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

Step-by-step 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`
`(2x)/(2)=(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`
`(-2y)/(-2)=-(x)/(-2)+(7)/(-2)`

Simplifying:

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

Removing the parentheses:

`=-8\cdot (-x+7)/(2)`

Multiplying fractions:

`=-(\left(-x+7\right)\cdot \:8)/(2)`
`\mathrm{Divide\:the\:numbers:}\:(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`
`(-2x)/(-2)=(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)