Solving Systems by Substitution: Additional Practice (Set 1)

Substitute $x = 2y + 2$ into $x + 4y = 14$.

$(2y + 2) + 4y = 14 \;\Rightarrow\; 6y + 2 = 14 \;\Rightarrow\; 6y = 12 \;\Rightarrow\; y = 2$

So $y = 2$.

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

$x = 2(2) + 2 = 6$

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

Back-substitute $w = 5$ into $l = w + 3$.

$l = 5 + 3 = 8$

So the length is $8$, and the perimeter $2(8) + 2(5) = 26$ checks out.

Substitute $y = 3x$ into $2x + 3y = 5$.

$2x + 3(3x) = 5 \;\Rightarrow\; 11x = 5 \;\Rightarrow\; x = \dfrac{5}{11}$

A fractional coordinate is perfectly valid.

Replace $x$ with $3 - 2y$ in $2x + 4y = 6$.

$2(3 - 2y) + 4y = 6 \;\Rightarrow\; 6 = 6 \text{ (always true)}$

An always-true statement means infinitely many solutions: one shared line.

Substitute $x = y + 4$ into $x + y = 12$.

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

So he has $4$ quarters.

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

$x = 4 + 4 = 8$

So there are $8$ dimes, and $8 + 4 = 12$ coins in all.

Back-substitute $x = 1$ into $y = 2 - 4x$.

$y = 2 - 4(1) = -2$

So the solution is $(1, -2)$.

Substitute $y = 2x - 1$ into $5x - 2y = 8$.

$5x - 2(2x - 1) = 8 \;\Rightarrow\; x + 2 = 8 \;\Rightarrow\; x = 6$

So $x = 6$.

Back-substitute $x = 6$ into $y = 2x - 1$.

$y = 2(6) - 1 = 11$

So $y = 11$, and the solution is $(6, 11)$.

Back-substitute $x = 1$ into $y = -3x$.

$y = -3(1) = -3$

So $y = -3$, and the solution is $(1, -3)$.

Replace $y$ in $4x - 2y = 7$ with $2x - 3$.

$4x - 2(2x - 3) = 7 \;\Rightarrow\; 6 = 7 \text{ (false)}$

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