Pearson Algebra 1 Common Core, 2011
PA
Pearson Algebra 1 Common Core, 2011 View details
6. Systems of Linear Inequalities
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Exercise 39 Page 405

Substitute the given point into the given systems.

A

Practice makes perfect

The given point (- 3,11) is a solution to a system of inequalities if it satisfies every inequality in the system. Let's substitute the point into the given systems one at a time until we find the correct one.

Option A

If the given point satisfies both inequalities at the same time, the point is a solution to the system. Let's substitute the point into both inequalities of the system.

y≥ x-2 & (I) 2x+y≤ 5 & (II)

(I), (II): x= -3, y= 11

11 ? ≥ -3 -2 2( -3)+ 11? ≤ 5
11 ? ≥- 3 -2 - 6 + 11? ≤5

(I), (II): Add and subtract terms

11 ≥ -5 ✓ 5≥ 5 ✓

Because we obtained true statements from both inequalities by substituting the point (-3,11), it is a solution to the system in option A.

Checking the Remaining Options

Although option A is correct, let’s verify our answer by checking the remaining options as well. To do so, we will following the same procedure for each of the remaining systems.

System Substitution Simplify
y > x+8 3x+y > 2 11 > - 3+8 3( -3)+ 11 > 2 11 > 5 ✓ 2 > 2 *
y > - x+8 2x+3y ≥ 7 11 > - ( -3)+8 2( -3)+3( 11) ≥ 7 11 > 11 * 27 ≥ 7 ✓
y≤ -3x+1 x-y≥ -15 11≤ -3( -3)+1 -3- 11≥ -15 11≤ 10 * - 14≥ -15 ✓

When we substitute the point (- 3, 11) into any of the remaining systems, at least one of the inequalities is not satisfied. Therefore, the point (-3,11) is not a solution of any of the remaining systems. This means that, as we discovered at the beginning of our solution, the correct answer is indeed option A.