Equations with Two Variables: Additional Practice (Set 2)

Substitute $x = 1$ and $y = \tfrac{5}{2}$ into each equation and look for a true statement.

$3(1) + 2\left(\tfrac{5}{2}\right) = 3 + 5 = 8$

So the pair satisfies $3x + 2y = 8$.

The solution of the system satisfies both equations, so test each pair in both.

$4 + 5 = 9 \quad\text{and}\quad 2(4) - 5 = 3$

So $(4, 5)$ works in both. Each other pair adds to $9$ but fails the second equation.

Substitute $x = 2$ into $2x + 3y = 12$ and solve, even though the answer is not a whole number.

$4 + 3y = 12, \qquad 3y = 8, \qquad y = \frac{8}{3}$

A fractional solution is still a valid solution.

A solution with $x = 0$ makes the $2x$ term vanish. Substitute and solve.

$2(0) + 5y = 20, \qquad 5y = 20, \qquad y = 4$

So the point is $(0, 4)$.

A point is on the line when its coordinates satisfy $2x + y = 6$, and coordinates need not be whole numbers.

$2(2.5) + 1 = 5 + 1 = 6$

So $(2.5, 1)$ is on the line. The others give $5$, $5$, and $7$.

Solve for $y$: $y = 10 - 3x$. Lowering $x$ by $1$ removes $-3$ from the right side, which adds $3$.

$y = 10 - 3x$

So a decrease of $1$ in $x$ makes $y$ increase by $3$.

Both points must satisfy the equation, so test each in the candidate.

$2(0) + 4 = 4 \quad\text{and}\quad 2(2) + 0 = 4$

Only $2x + y = 4$ holds for both points.

A point is on the line when $3x - y = 5$, so the exception is the pair that fails it.

$3(1) - 1 = 2 \neq 5$

So $(1, 1)$ is not on the line. The other three each give $5$.

Expand the parentheses first, then move the constant across.

$3(x - 2) + y = 0 \;\Rightarrow\; 3x - 6 + y = 0 \;\Rightarrow\; 3x + y = 6$

List the whole-number pairs with $x \ge 0$ and $y \ge 0$ that add to $3$.

$(0, 3), \; (1, 2), \; (2, 1), \; (3, 0)$

That is $4$ pairs. The full line has infinitely many solutions, but this restriction leaves only these four.

Two adult tickets at $6$ dollars cost $12$ dollars, so the three child tickets make up the rest of the $27$.

$12 + 3y = 27, \qquad 3y = 15, \qquad y = 5$

Raising $x$ by $1$ adds $2$ to the $2x$ term.

$2(x + 1) + 1 = (2x + 1) + 2$

So $y$ increases by $2$ each time.