Question

#1: Solve the linear system below using the elimination method. Type your

answer as an ordered pair in the form (#,#).*
5x + 3y = 1
-5x – 7y = 31

Answer 1

The given linear system is:

`\displaystyle \left \{ {{5x+3y=1} \atop {-5x-7y=31}} \right.`

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.

`5x+3y=1\\-5x-7y=31`

`\displaystyle 3y-7y=1+31\\-4y=32`

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:

`\displaystyle(-4y)/(-4) =(32)/(-4)`

`\displaystyle y=-8`

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.

`\displaystyle 5x+3y=1 \rightarrow 5x+3(-8)=1`

Open the parentheses and multiply:

`5x-24=1`

Move 24 to the other side to leave the variable alone:

`5x-24+24=1+24`

You will be adding since you're "removing" it by doing the opposite of it.

`5x=25`

Lastly, divide both sides by 5 to leave x alone.

`\displaystyle (5x)/(5) =(25)/(5)`

`x=5`

`\displaystyle (x,y) \rightarrow (5, -8)`

The answer is (5, -8).