Solving Systems by Elimination: Practice

Back-substitute $x = 2$ into $x + 2y = 8$.

$2 + 2y = 8 \;\Rightarrow\; 2y = 6 \;\Rightarrow\; y = 3$

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

The first has $y$, the second has $3y$. To turn the first's $y$ into $-3y$, multiply the whole first equation by $-3$.

$-3(2x + y = 7) \;\Rightarrow\; -6x - 3y = -21$

Adding this to $x + 3y = 11$ then cancels $y$.

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

$3 - y = 1 \;\Rightarrow\; y = 2$

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

Multiply the first by $3$ so both equations carry $3y$.

$3(4x + y = 10) \;\Rightarrow\; 12x + 3y = 30$

Subtract $x + 3y = 8$: $11x = 22$, so $x = 2$.

Back-substitute $x = 2$ into $4x + y = 10$.

$4(2) + y = 10 \;\Rightarrow\; y = 2$

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

Multiply the second by $-2$: $-2x - 2y = -4$. Add it to $2x + 3y = 1$ so the $x$ terms cancel.

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

So $y = -3$.

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

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

So the solution pair is $(5, -3)$.

Both equations carry $x$, so subtracting removes it.

$(x + 4y) - (x + y) = 14 - 5 \;\Rightarrow\; 3y = 9 \;\Rightarrow\; y = 3$

So $y = 3$.

Back-substitute $y = 3$ into $x + y = 5$.

$x + 3 = 5 \;\Rightarrow\; x = 2$

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

Both variables cancel and a false statement is left.

$0 = 4 \text{ is false} \;\Rightarrow\; \text{no solution}$

The two lines are parallel.

The $y$ terms are $-2y$ and $+2y$, opposites, so adding cancels $y$.

$(5x - 2y) + (3x + 2y) = 4 + 12 \;\Rightarrow\; 8x = 16$

So adding is the immediate move.

An always-true statement means every solution of one equation solves the other.

$0 = 0 \;\Rightarrow\; \text{same line, infinitely many solutions}$

The two equations are different forms of one line.