Solving Systems by Substitution: Practice

Replace $y$ in $2x + y = 7$ with $x + 1$.

$2x + (x + 1) = 7 \;\Rightarrow\; 3x + 1 = 7 \;\Rightarrow\; 3x = 6 \;\Rightarrow\; x = 2$

So $x = 2$.

Replace $y$ in $x + y = 5$ with $x - 1$.

$x + (x - 1) = 5 \;\Rightarrow\; 2x - 1 = 5 \;\Rightarrow\; x = 3$

Then $y = x - 1 = 2$, so the solution is $(3, 2)$.

The second equation already gives $x = 3$, so put $x = 3$ into $x + 2y = 7$.

$3 + 2y = 7 \;\Rightarrow\; 2y = 4 \;\Rightarrow\; y = 2$

So $y = 2$.

Both right-hand sides equal $y$, so they equal each other.

$2x - 3 = x + 1$

This single equation has only $x$ in it and can be solved directly.

Subtract $x$ from both sides, then add $3$.

$2x - 3 = x + 1 \;\Rightarrow\; x - 3 = 1 \;\Rightarrow\; x = 4$

So $x = 4$.

Replace $y$ in $3x + y = 12$ with $x + 4$.

$3x + (x + 4) = 12 \;\Rightarrow\; 4x + 4 = 12 \;\Rightarrow\; 4x = 8 \;\Rightarrow\; x = 2$

So $x = 2$.

Back-substitute $x = 2$ into $y = x + 4$.

$y = 2 + 4 = 6$

So $y = 6$, and the solution pair is $(2, 6)$.

Only $3y$ is added to $x$, so subtract $3y$ from both sides.

$x = 9 - 3y$

The $3y$ crosses the equals sign with the opposite sign, becoming $-3y$.

Replace $x$ in $2x - y = 4$ with the whole expression $9 - 3y$, keeping it in parentheses.

$2(9 - 3y) - y = 4$

The $2$ multiplies the entire expression, and the $-y$ term stays.

Distribute the $2$, then collect the $y$ terms.

$18 - 6y - y = 4 \;\Rightarrow\; 18 - 7y = 4 \;\Rightarrow\; -7y = -14 \;\Rightarrow\; y = 2$

So $y = 2$.

Replace $y$ in $2x + y = 8$ with $5 - x$.

$2x + (5 - x) = 8 \;\Rightarrow\; x + 5 = 8 \;\Rightarrow\; x = 3$

So $x = 3$.

Both right-hand sides equal $y$, so set them equal.

$2x + 1 = 3x - 2 \;\Rightarrow\; 1 + 2 = 3x - 2x \;\Rightarrow\; x = 3$

So $x = 3$.