Hey there!
When we use substitution, we take one of our equations and solve for a variable there- then use that solved expression and plug it into the other equation to solve for the other variable. Then, we use that solved variable and solve for the original. Let's try that here.
We'll use x - 4y = 2 and solve for x. We add 4y to both sides to get:
x = 4y + 2
Now, we plug 4y + 2 in for x to solve for y in the other equation:
3(4y + 2) + 5y = 40
Use the distributive property which states that a(b+c) = ab + ac
12y + 6 + 5y = 40
17y + 6 = 40
17y = 34
y = 2
Now, we can use that value of 2 and plug it back into the first equation to solve for x.
x -4(2) = 2
x - 8 = 2
x = 10
To check if this is true, we can plug in 10 for x and 2 for y into 3x + 5y = 40 and see if it works.
3(10) + 5(2) = 40
30 + 10 = 40
40 = 40
Therefore, x = 10 and y = 2.
Hope this helps!