The exercise tells us that Aimee thinks the solution to the system shown below is
(-4,-6), and Eric thinks the solution is
(8,2). We need to find who is correct and explain our reasoning.
{2x−3y=106y=4x−20
Recall that for a point to be the solution of a system of equations it needs to satisfy both equations. Let's check whose solution gives a true statement when substituting it in both equations. We will check each of them individually.
Aimee's Solution
We will check Aimee's solution first. We will start by substituting
(-4,-6) in the first equation.
2x−3y=10
2(-4)−3(-6)=?10
-8+18=?10
10=10
We can see that Aimee's solutions satisfy the first equation. Let's check if it also satisfies the second one.
6y=4x−20
6(-6)=?4(-4)−20
-36=?-16−20
-36=-36
Once more we obtained a true statement. Aimee's solution satisfies both equations, and therefore her solution is correct.
Eric
We will start by substituting Eric's solution,
(8,2), in the first equation.
2x−3y=10
2(8)−3(2)=?10
16−6=?10
10=10
Eric's solutions satisfies the first equation. Let's now check to see if it also works for the second equation.
6y=4x−20
6(2)=?4(8)−20
12=?32−20
12=12
As Eric's solution satisfies both equations, he is correct as well.
Conclusions
After checking both Aimee and Eric's solutions, we can see that both are correct. How can this be?
If we manipulate one of the equations from our system following the Properties of Equality, we can find our answer.
{2x−3y=106y=4x−20(I)(II)
{2(2x−3y=10)6y=4x−20
{2⋅2x−2⋅3y=2⋅106y=4x−20
{4x−6y=206y=4x−20
{4x−6y+6y=20+6y6y=4x−20
{4x=20+6y6y=4x−20
{4x−20=20+6y−206y=4x−20
{4x−20=6y6y=4x−20
{6y=4x−206y=4x−20
As we can see, both equations are equivalent. Therefore there will be infinitely many solutions for the system. Every point satisfying
6y=4x−20 will be a solution to the system.