Here is your system of equations:
[tex]\left \{ {{y=-4x+10} \atop {y=-2x-6}} \right.[/tex]
When solving a system, you can either substitute, or eliminate. I am going to substitute.
In the first equation, if y is equal to -4x + 10, I can substitute the y in the second equation with -4x + 10.
Here is your new equation:
[tex]-4x+10=-2x-6[/tex]
Since you have variables on both sides, you have to cancel it on either side. In the equation on the right, the variable is -2x. To cancel it out, you have to do the opposite of it, which is +2x, or positive 2x. That means you have to add 2x to both equations.
[tex]-4x+10+2x = -2x-6+2x \\ \\ -2x+10=-6[/tex]
Now you have one variable and two regular integers. You have to leave the variable x alone to find its value. Along with it is +10, or positive 10. To remove it, you have to do the opposite of it, which is -10 or subtract 10.
[tex]-2x+10-10=-6-10 \\ \\ -2x=-16[/tex]
- Remember that when you subtract from a negative number, the number will increase but the sign will stay negative.
Now you need x to stay alone, so divide both sides by -2.
[tex]\frac{-2x}{-2}=\frac{-16}{-2} \\ \downarrow \\ x=8[/tex]
Remember:
NEGATIVES
- Negative(-) times(×) negative(-) = positive(+)
- Negative(-) times(×) positive(+) = negative(-)
- Negative(-) divided(÷) by positive(+) = negative(-)
- Negative(-) divided(÷) by negative(-) = positive(+)
POSITIVES
- Positive(+) times(×) positive(+) = positive(+)
- Positive(-) times(×) negative(-) = negative(-)
- Positive(+) divided(÷) by positive(+) = positive(+)
- Positive(+) divided(÷) by negative(-) = negative(-)
Now that you have the value of x, you can substitute it with the x in one of the equations.
[tex]y=-4x+10 \rightarrow y=-4(8)+10[/tex]
Multiply -4 by 8:
[tex]-4 \times 8 = -32[/tex]
Here is your new equation:
[tex]y=-32+10[/tex]
Add -22 and 10:
[tex]-32+10 =-22[/tex]
[tex]\left \{ {{x=8} \atop {y=-22}} \right.[/tex]
Your answer is the third choice: (8, -22)
If you have any questions, feel free to ask in the comments! :)
