Question

Is (10,5)
a viable solution for the following system? Explain.

x+y<25

12x+20y≥250

(1 point)
Responses

(10,5)
is a nonviable solution since it does not make both inequalities true.
left parenthesis 10 comma 5 right parenthesis is a nonviable solution since it does not make both inequalities true.

(10,5)
is a nonviable solution since it makes both inequalities true.
left parenthesis 10 comma 5 right parenthesis is a nonviable solution since it makes both inequalities true.

(10,5)
is a viable solution since it makes one inequality true.
left parenthesis 10 comma 5 right parenthesis is a viable solution since it makes one inequality true.

(10,5)
is a viable solution since it makes both inequalities true.
left parenthesis 10 comma 5 right parenthesis is a viable solution since it makes both inequalities true.
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Answer 1

To determine if (10, 5) is a viable solution to the system of inequalities, we need to evaluate both inequalities with the values of \( x = 10 \) and \( y = 5 \).

1. Check the first inequality:
\[
x + y < 25
\]
Substituting \( x = 10 \) and \( y = 5 \):
\[
10 + 5 < 25 \\
15 < 25 \quad \text{(true)}
\]

2. Check the second inequality:
\[
12x + 20y \geq 250
\]
Substituting \( x = 10 \) and \( y = 5 \):
\[
12(10) + 20(5) \geq 250 \\
120 + 100 \geq 250 \\
220 \geq 250 \quad \text{(false)}
\]

Since (10, 5) makes the first inequality true but does **not** satisfy the second inequality, it does not meet the requirements of both inequalities.

Thus, the correct response is:

**(10,5) is a nonviable solution since it does not make both inequalities true.**