Solving Systems by Substitution: Practice

Back-substitute $x = 4$ into $y = 11 - 2x$.

$y = 11 - 2(4) = 3$

So $y = 3$, and the solution is $(4, 3)$.

Substitute $y = 2x - 4$ into $3x + 2y = 6$.

$3x + 2(2x - 4) = 6 \;\Rightarrow\; 7x - 8 = 6 \;\Rightarrow\; 7x = 14 \;\Rightarrow\; x = 2$

So $x = 2$.

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

$y = 2(2) - 4 = 0$

So the solution is $(2, 0)$.

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

$x = 3(1) + 1 = 4$

So $x = 4$, and the solution is $(4, 1)$.

Replace $y$ in $x + y = 10$ with $x + 2$.

$x + (x + 2) = 10 \;\Rightarrow\; 2x + 2 = 10 \;\Rightarrow\; 2x = 8 \;\Rightarrow\; x = 4$

So $x = 4$.

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

$y = 4 + 2 = 6$

So the solution is $(4, 6)$.

In $2x + y = 7$ the variable $y$ has coefficient $1$, so subtract $2x$ from both sides.

$y = 7 - 2x$

The $2x$ crosses the equals sign with the opposite sign.

Replace $y$ with $7 - 2x$ and distribute the $3$.

$4x + 3(7 - 2x) = 17 \;\Rightarrow\; -2x + 21 = 17 \;\Rightarrow\; -2x = -4 \;\Rightarrow\; x = 2$

So $x = 2$.

Back-substitute $x = 2$ into $y = 7 - 2x$.

$y = 7 - 2(2) = 3$

So the solution is $(2, 3)$.

Replace $y$ in $x + y = 20$ with $x + 4$.

$x + (x + 4) = 20 \;\Rightarrow\; 2x + 4 = 20 \;\Rightarrow\; 2x = 16 \;\Rightarrow\; x = 8$

So the smaller number is $8$.

Back-substitute $x = 3$ into $y = -x + 6$.

$y = -(3) + 6 = 3$

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

An always-true statement means every pair on the first line also satisfies the second.

$4 = 4 \;\Rightarrow\; \text{same line, infinitely many solutions}$

The two equations are just different forms of one line.