Question

Solve each system using substitution show your answer as an ordered pair no solution or infinite solution

Answer 1

1) Given y = x+2 and y = -4x-8.

Since the left hand sides of both equations are same, equate the right hand side of both the equations.

`x+2=-4x-8`

Add 4x on both sides.

`\begin{gathered} x+2+4x=-4x-8+4x \\ 5x+2=-8 \end{gathered}`

Add -2 on both sides.

`\begin{gathered} 5x+2-2=-8-2 \\ 5x=-10 \end{gathered}`

Divide by 5 on both sides.

`\begin{gathered} x=-(10)/(5) \\ =-2 \end{gathered}`

Substitute the value of x into y = x+2.

`\begin{gathered} y=-2+2 \\ =0 \end{gathered}`

Solution is (-2,0).

2) Given y = 3x+1 and y = -2x+6.

Since the left hand sides of both equations are same, equate the right hand side of both the equations.

`3x+1=-2x+6`

Add 2x on both sides.

`\begin{gathered} 3x+1+2x=-2x+6+2x \\ 5x+1=6 \end{gathered}`

Add -1 on both sides.

`\begin{gathered} 5x+1-1=6-1 \\ 5x=5 \end{gathered}`

Divide by 5 on both sides.

`\begin{gathered} x=(5)/(5) \\ =1 \end{gathered}`

Substitute the value of x into y = 3x+1.

`\begin{gathered} y=3\cdot1+1 \\ =4 \end{gathered}`

Solution is (1, 4).

3) Given y = -3x-6 and 6x+2y = -2.

Substitute -3x-6 for y into 6x+2y = -2.

`\begin{gathered} 6x+2(-3x-6)=-2 \\ 6x-6x-12=-2 \\ -12=-2 \end{gathered}`

which is not possible. Hence the given system of equations has no solution.

4) Given y = -5 and -8x+4=-20.

From the second equation, -8x+4 = -20, solve for x.

Add -4 on both sides.

`\begin{gathered} -8x+4-4=-20-4 \\ -8x=-24 \end{gathered}`

Divide by -8 on both sides.

`\begin{gathered} x=(-24)/(-8) \\ =3 \end{gathered}`

Solution is (-5, 3).