Solving Systems by Substitution: Practice

Substitute $x = y + 2$ into $4x - 3y = 5$.

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

So $y = -3$.

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

$x = -3 + 2 = -1$

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

Back-substitute $x = 1$ into $y = 5 - 2x$.

$y = 5 - 2(1) = 3$

So the solution is $(1, 3)$.

Substitute $y = 2x$ into $3x + 2y = 10$.

$3x + 2(2x) = 10 \;\Rightarrow\; 7x = 10 \;\Rightarrow\; x = \dfrac{10}{7}$

A fraction is a perfectly valid coordinate.

Distribute and simplify.

$4x - 4x + 8 = 8 \;\Rightarrow\; 8 = 8 \text{ (always true)}$

An always-true statement means infinitely many solutions: the two equations are the same line.

Replace $y$ in $x + y = 7$ with $x + 3$.

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

So the pen costs $2$ dollars.

Substitute $x = 2 - 2y$ into $3x + 4y = 10$.

$3(2 - 2y) + 4y = 10 \;\Rightarrow\; 6 - 2y = 10 \;\Rightarrow\; -2y = 4 \;\Rightarrow\; y = -2$

So $y = -2$.

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

$x = 2 - 2(-2) = 2 + 4 = 6$

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

Replace $y$ in $x + y = 30$ with $2x$.

$x + 2x = 30 \;\Rightarrow\; 3x = 30 \;\Rightarrow\; x = 10$

Then $y = 2x = 20$, so the larger number is $20$.

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

$y = 2 - 3 = -1$

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

Substitute $x = 1 - y$ into $2x + 3y = 1$.

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

So $y = -1$.

Replace $y$ with $7x - 3$ and distribute the $5$.

$2x + 5(7x - 3) = 22 \;\Rightarrow\; 37x - 15 = 22 \;\Rightarrow\; 37x = 37 \;\Rightarrow\; x = 1$

So $x = 1$, and back-substituting gives $y = 4$.