1Cademy - Checking Whether $$(3, -1)$$ and $$(6, -3)$$ are Solutions of $$\{x

Activity (Process)

Checking Whether $(3, -1)$ and $(6, -3)$ are Solutions of $\{x - 5y > 10,\; 2x + 3y > -2\}$

Determine whether each ordered pair is a solution to the system $\left\{\begin{array}{l} x - 5y > 10 \\ 2x + 3y > -2 \end{array}\right.$

ⓐ Testing $(3, -1)$ : Substitute $x = 3$ and $y = -1$ into both inequalities.

First inequality: $3 - 5(-1) = 3 + 5 = 8$ . Since $8 > 10$ is false, the first inequality is not satisfied.
Second inequality: $2(3) + 3(-1) = 6 - 3 = 3$ . Since $3 > -2$ is true, the second inequality is satisfied.

Because $(3, -1)$ fails to satisfy the first inequality, it is not a solution to the system.

ⓑ Testing $(6, -3)$ : Substitute $x = 6$ and $y = -3$ into both inequalities.

First inequality: $6 - 5(-3) = 6 + 15 = 21$ . Since $21 > 10$ is true, the first inequality is satisfied.
Second inequality: $2(6) + 3(-3) = 12 - 9 = 3$ . Since $3 > -2$ is true, the second inequality is also satisfied.

Because $(6, -3)$ makes both inequalities true, it is a solution to the system.

Updated 2026-04-28

Contributors are:

Who are from:

References