Problem 1 (on the left)
We can solve for y in the second equation since the coefficient here is 1.
Having coefficients that aren't -1 or 1 means we'd likely end up with fractions.
Add 5x to both sides like so:
y - 5x = 19
y = 5x+19
Now plug this into the first equation.
-3x - 8y = 20
-3x - 8( y ) = 20
-3x - 8( 5x+19 ) = 20
At this point, we have only one variable.
We stop here because your teacher does not want you to solve the final equation.
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Answers:
Variable to isolate: y
It leads to y = 5x+19 after solving the second equation for y.
And further leads to -3x - 8( 5x+19 ) = 20 after substitution.
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Problem 2 (on the right)
This time we can solve for x in the second equation. The reasoning is the same (coefficient of 1). Subtract 3y from both sides to get:
x+3y = 1
x = -3y+1
Substitute this into first equation.
Everywhere you see an x, replace it with -3y+1.
-3x - 3y = -15
-3( x ) - 3y = -15
-3( -3y+1 ) - 3y = -15
Like before, we now have one variable left.
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Answers:
Variable to isolate: x
It leads to x = -3y+1 when you solve for x in the second equation.
And further leads to -3( -3y+1 ) - 3y = -15 after substitution.
