Equations with Two Variables: Practice

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

$4 + 3 = 7 \quad\text{and}\quad 4 - 3 = 1$

So $(4, 3)$ works in both. Each other pair adds to $7$ but fails the second equation.

Substitute $x = 3$ into $4x + 6y = 24$ and solve.

$12 + 6y = 24, \qquad 6y = 12, \qquad y = 2$

Take half of the first coordinate, add the second, and look for $5$.

$\tfrac{1}{2}(4) + 3 = 2 + 3 = 5$

So $(4, 3)$ is a solution. The others give $4$, $6$, and $4$.

Find each unknown coordinate, then add. From $a + 6 = 10$ comes $a = 4$, and from $2 + b = 10$ comes $b = 8$.

$a + b = 4 + 8 = 12$

Substitute $x = 3$ into $2x - 3y = -3$ and solve.

$6 - 3y = -3, \qquad -3y = -9, \qquad y = 3$

A point lies on the line when $x - 2y = 4$. Subtract twice the second coordinate from the first.

$6 - 2(1) = 6 - 2 = 4$

So $(6, 1)$ is on the line. The others give $0$, $0$, and $1$.

Solve for $y$: $y = 12 - x$. Raising $x$ by $1$ subtracts $1$ from the right side.

$y = 12 - x$

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

Multiply the first coordinate by $5$, subtract twice the second, and look for $1$.

$5(1) - 2(2) = 5 - 4 = 1$

So $(1, 2)$ is a solution. The others give $8$, $11$, and $3$.

Substitute $x = 4$ and $y = -1$ into each equation and find the true one.

$2(4) + 3(-1) = 8 - 3 = 5$

So $(4, -1)$ satisfies $2x + 3y = 5$.

Put $x = k$ and $y = k$ into $3x + 2y = 20$ and combine like terms.

$3k + 2k = 20, \qquad 5k = 20, \qquad k = 4$

The line contains both points, so both must satisfy the equation. Test each in the candidate.

$3(1) - 2 = 1 \quad\text{and}\quad 3(3) - 8 = 1$

Only $3x - y = 1$ holds for both points.

For every value of $x$, the equation gives one matching $y = 100 - x$.

$y = 100 - x$

Since $x$ can be any of infinitely many numbers, there are infinitely many pairs.