The given linear system is:
[tex]\displaystyle \left \{ {{5x+3y=1} \atop {-5x-7y=31}} \right.[/tex]
Linear systems can be solved using either elimination or substitution. However, the question is asking to solve using elimination, so I will use that method.
When eliminating, you can either eliminate x or y. In this system, x is much easier to eliminate. The x variable in the first equation is 5x, and in the second equation, it is -5x. Since 5 and -5 cancel each other out, you don't need to do anything other than add.
[tex]5x+3y=1\\-5x-7y=31[/tex]
[tex]\displaystyle 3y-7y=1+31\\-4y=32[/tex]
Lastly, you need to leave the variable y alone. The variable is currently -4y or -4 times y. To remove it, you need to do the opposite of it, which is dividing by -4.
Divide both sides by -4:
[tex]\displaystyle\frac{-4y}{-4} =\frac{32}{-4}[/tex]
[tex]\displaystyle y=-8[/tex]
Now that you have the value of y, substitute it into one of the equations to find x. I will be substituting it into the first equation.
[tex]\displaystyle 5x+3y=1 \rightarrow 5x+3(-8)=1[/tex]
Open the parentheses and multiply:
[tex]5x-24=1[/tex]
Move 24 to the other side to leave the variable alone:
[tex]5x-24+24=1+24[/tex]
You will be adding since you're "removing" it by doing the opposite of it.
[tex]5x=25[/tex]
Lastly, divide both sides by 5 to leave x alone.
[tex]\displaystyle \frac{5x}{5} =\frac{25}{5}[/tex]
[tex]x=5[/tex]
[tex]\displaystyle (x,y) \rightarrow (5, -8)[/tex]
The answer is (5, -8).
