A point in the coordinate plane must satisfy both of the following conditions: 1. 2. What is the greatest integer value of for which such a point can exist?

Solution available with step-by-step explanation

SAT Linear Inequalities Question 90 | Free SAT Prep | Aniko

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Question 90·Hard·Linear Inequalities in One or Two Variables

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A point $(x,y)$ in the coordinate plane must satisfy both of the following conditions:

$2x + 3y \le 15$
$y \ge x + 4$

What is the greatest integer value of $x$ for which such a point can exist?

Your answer

(Express the answer as an integer)

For SAT questions about regions defined by linear inequalities that ask for a maximum or minimum value of a variable, first interpret each inequality as a line plus a shading direction. The extreme (maximum or minimum) values usually occur on the boundary where the relevant lines intersect, so set the corresponding equations equal, solve for the intersection, and then apply any extra condition (like “integer” or “positive”) to that value. Always check whether you should round up, round down, or use the exact value based on the inequality signs.

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Visualize or think about the two inequalities

Try to think of (or sketch) the two lines $2x + 3y = 15$ and $y = x + 4$ . One inequality wants points below a line, and the other wants points above another line. Where can they both be true at once?

Focus on the boundary where both are just satisfied

To find the largest possible $x$ that still works, consider the point where both inequalities are equalities: $2x + 3y = 15$ and $y = x + 4$ . What happens if you substitute $y = x + 4$ into the first equation?

Use the intersection to pick an integer

After you find the $x$ -coordinate where the lines intersect, remember the question asks for an integer $x$ . Think about which integers are less than or equal to that $x$ -value, and choose the greatest one.

Desmos Guide

Graph the boundary lines

In Desmos, enter 2x + 3y = 15 on one line and y = x + 4 on another line so you can see both boundary lines on the same coordinate plane.

Find their intersection point

Click on the point where the two lines intersect; Desmos will display the coordinates of this intersection. Focus on the $x$ -coordinate of this point.

Determine the greatest integer from the intersection $x$ -value

Look at the $x$ -coordinate you found and identify the greatest integer that is less than or equal to this value; that integer is the greatest $x$ for which a point can satisfy both inequalities.

Step-by-step Explanation

Understand the region described by each inequality

Rewrite each inequality as a line plus a shading rule:

$2x + 3y \le 15$ describes all points on or below the line $2x + 3y = 15$ .
$y \ge x + 4$ describes all points on or above the line $y = x + 4$ .

Any point $(x, y)$ that works must lie in the overlap of these two regions.

Find where the boundary lines meet

To get the largest possible $x$ that still satisfies both inequalities, the point will sit right on both boundary lines at the same time.

So set $y = x + 4$ and plug into $2x + 3y = 15$ (using equality because we are on the boundary):

2x + 3(x + 4) = 15

Now simplify this equation.

Solve for the $x$ -coordinate of the intersection

Continue solving:

2x + 3(x + 4) = 15 \\[0.7em] 2x + 3x + 12 = 15 \\[0.7em] 5x + 12 = 15 \\[0.7em] 5x = 3 \\[0.7em] x = \frac{3}{5}

So the two lines intersect at $x = \frac{3}{5}$ (and some $y$ value). This is the largest real $x$ for which a point can satisfy both conditions.

Choose the greatest integer $x$ that works

The question asks for the greatest integer $x$ that still allows a point to satisfy both inequalities.

Since any valid $x$ must be less than or equal to $\frac{3}{5}$ , the greatest integer that is $\le \frac{3}{5}$ is $0$ .

So, the greatest integer value of $x$ for which such a point can exist is $0$ .

Strategy

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation