Question

5. Solve the 2. Solve by substitution. x - 2y = -2 A. (-10-5 A. (0.2 B. (9) B. (-2,0) C. no D. inf D. (0,0) 6. Solved y 3. Solve by elimination.

Answer 1

`\begin{cases}x-2y=-2 \\ y=2x+4\end{cases}`

We are given the system of equations. We need to solve by substitution.

The second equation is already solved for y, so we will use this equation and substitute it in place for the y-variable in the first equation.

`\begin{gathered} x-2(2x+4)=-2 \\ x-4x-8=-2 \\ -3x-8=-2 \\ -3x=6 \\ x=-2 \end{gathered}`

We will use the distributive property in the beginning to distribute the -2 into the 2x and the 4:

`\begin{gathered} -2\cdot2x=-4x \\ -2\cdot4=-8 \end{gathered}`

This will update our equation to:

`x-4x-8=-2`

Then, we will combine like terms, first with our constants (no variables):

`\begin{gathered} x-4x-8=-2 \\ x-4x=6 \end{gathered}`

We add 8 to both sides of the equation to remove it from the left side and place it on the right. This way, we are one step closer to isolating our x-variable. Now, combine the variable terms.

`\begin{gathered} x-4x=6 \\ -3x=6 \end{gathered}`

We imply a 1 in front of the lone x, so 1 - 4 = -3. Therefore, our coefficient for our variable becomes -3. Finally, we will divide both sides by -3 to fully isolate our x-variable.

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

Now, we will use our value for x and plug it into either given equation. I will use the second given equation in the question.

`\begin{gathered} y=2x+4 \\ y=2(-2)+4 \\ y=-4+4 \\ y=0 \end{gathered}`

We first substitute our x-value, -2, into the place of x in the equation.

`y=2(-2)+4`

Then, we multiply 2 and -2.

`y=-4+4`

Finally, we add -4 to 4.

`y=0`

Now, we are going to check our work. We will substitute the values into both equations and make sure we get a true statement.

Let's use equation 1 first: x - 2y = -

`\begin{gathered} x-2y=-2 \\ -2-2(0)=-2 \\ -2-0=-2 \\ -2=-2 \end{gathered}`