Solving Systems by Substitution: Additional Practice (Set 2)

Replace $y$ in $4x - y = 1$ with $5 - 2x$, keeping the parentheses.

$4x - (5 - 2x) = 1 \;\Rightarrow\; 6x - 5 = 1 \;\Rightarrow\; 6x = 6 \;\Rightarrow\; x = 1$

So $x = 1$.

Substitute $l = w + 3$ into $2l + 2w = 26$.

$2(w + 3) + 2w = 26 \;\Rightarrow\; 4w + 6 = 26 \;\Rightarrow\; 4w = 20 \;\Rightarrow\; w = 5$

So the width is $5$.

Distribute and watch the variable terms cancel.

$6x - 6x + 4 = 5 \;\Rightarrow\; 4 = 5 \text{ (false)}$

A false statement means no solution: the lines are parallel.

Replace $y$ with $-2x + 4$ in $4x + 2y = 3$.

$4x + 2(-2x + 4) = 3 \;\Rightarrow\; 8 = 3 \text{ (false)}$

A false statement means no solution: the lines are parallel.

Back-substitute $x = 2$ into $y = x + 3$.

$y = 2 + 3 = 5$

So the notebook costs $5$ dollars, and $2 + 5 = 7$ checks out.

Substitute $x = 2y$ into $x + y = 18$.

$2y + y = 18 \;\Rightarrow\; 3y = 18 \;\Rightarrow\; y = 6$

So the brother is $6$, and Maria is $12$.

Replace $x$ with $6 - y$ in the equation it came from.

$(6 - y) + y = 6 \;\Rightarrow\; 6 = 6$

The variable cancels and the statement is always true, so it tells you nothing new. Always substitute into the other equation.

Replace $y$ in $5x - 2y = 11$ with $-3x$.

$5x - 2(-3x) = 11 \;\Rightarrow\; 11x = 11 \;\Rightarrow\; x = 1$

So $x = 1$.

Solve the first for $x = 2y - 1$ and substitute into $2x + y = 8$.

$2(2y - 1) + y = 8 \;\Rightarrow\; 5y - 2 = 8 \;\Rightarrow\; y = 2$

Then $x = 2(2) - 1 = 3$, so $(3, 2)$ solves both.

Substitute $x = y + 4$ into $x + 2y = 19$.

$(y + 4) + 2y = 19 \;\Rightarrow\; 3y + 4 = 19 \;\Rightarrow\; 3y = 15 \;\Rightarrow\; y = 5$

So a child ticket costs $5$ dollars.

Look for a coefficient of $1$ or $-1$. In $7x - y = 3$ the variable $y$ has coefficient $-1$, so isolating it avoids fractions.

$y = 7x - 3$

Every other choice divides by $2$, $5$, or $7$.

Back-substitute $y = 5$ into $x = y + 4$.

$x = 5 + 4 = 9$

So an adult ticket costs $9$ dollars, and $9 + 2(5) = 19$ checks out.