For what value of does the system of equations have a solution in which both and are integers?

Solution available with step-by-step explanation

SAT Systems of Two Linear Equations Question 33 | Free SAT Prep | Aniko

00:00

Question 33·Hard·Systems of Two Linear Equations in Two Variables

0 / 140

For what value of $k$ does the system of equations

(k+3)x + 2y = 19

5x + (k - 1)y = 22

have a solution $(x, y)$ in which both $x$ and $y$ are integers?

For systems with a parameter and multiple-choice options, it is usually fastest to plug each choice for the parameter into the equations and solve the resulting system using elimination. For each candidate value of $k$ , eliminate one variable quickly to see if $x$ (or $y$ ) becomes a fraction—if it does, discard that $k$ and move on. As soon as you find a $k$ that produces integer values for both $x$ and $y$ , you can stop and choose that answer without doing any extra algebra.

Hints

Click me!

Use the answer choices for k

Notice that the only unknown parameter is $k$ , and there are only four possible values. Instead of solving symbolically for $k$ , try substituting each answer choice into the system and solving for $x$ and $y$ .

Solve each resulting system efficiently

When you plug in a value for $k$ , you get a regular $2$ -equation system in $x$ and $y$ . Use elimination (add or subtract the equations, or multiply one equation) to eliminate a variable quickly and see whether $x$ or $y$ comes out as a fraction or an integer.

Check the integrality condition

You do not need the exact values for all four choices. As soon as you find a value of $k$ that gives both $x$ and $y$ as integers, you can stop and choose that option.

Desmos Guide

Enter the parametric system

In one expression line, type (k+3)x + 2y = 19. In another line, type 5x + (k-1)y = 22. Desmos will graph both equations as curves in the $xy$ -plane once $k$ has a value.

Create and use a slider for k

Add a new line k = 0 to create a slider for $k$ . Then adjust the slider (or type directly) to set $k$ equal to each answer choice: $-1$ , $0$ , $2$ , and $4$ in turn.

Find and inspect the intersection point for each k

For each value of $k$ , click on the point where the two graphs intersect; Desmos will display its coordinates $(x,y)$ . Check whether both the $x$ - and $y$ -coordinates are integers.

Choose the k that gives integer coordinates

Among the tested $k$ -values, identify the one for which the intersection point has integer $x$ and integer $y$ . That corresponding $k$ is the value you should select on the test.

Step-by-step Explanation

Plan: Test each value of k from the answer choices

The equations are

\begin{aligned} (k+3)x + 2y &= 19 \\ 5x + (k - 1)y &= 22 \end{aligned}

We only have four possible values for $k$ ( $-1, 0, 2, 4$ ), so it is fastest to plug in each value of $k$ , solve the resulting system for $x$ and $y$ , and see whether both are integers.

Test k = -1 and k = 0

For $k=-1$ :

Then $k+3=2$ and $k-1=-2$ , so the system becomes

\begin{aligned} 2x + 2y &= 19 \\ 5x - 2y &= 22 \end{aligned}

Add the equations:

\begin{aligned} (2x+2y) + (5x-2y) &= 19 + 22 \\ 7x &= 41 \\ x &= \frac{41}{7} \end{aligned}

Since $x$ is not an integer, $k=-1$ does not work.

For $k=0$ :

Then $k+3=3$ and $k-1=-1$ , so the system becomes

\begin{aligned} 3x + 2y &= 19 \\ 5x - y &= 22 \end{aligned}

Multiply the second equation by $2$ and add:

\begin{aligned} 3x + 2y &= 19 \\ 10x - 2y &= 44 \end{aligned}

Add them:

\begin{aligned} (3x+2y) + (10x-2y) &= 19 + 44 \\ 13x &= 63 \\ x &= \frac{63}{13} \end{aligned}

Again, $x$ is not an integer, so $k=0$ does not work either.

Test k = 2 and k = 4

For $k=2$ :

Then $k+3=5$ and $k-1=1$ , so the system becomes

\begin{aligned} 5x + 2y &= 19 \\ 5x + y &= 22 \end{aligned}

Subtract the second equation from the first:

\begin{aligned} (5x+2y) - (5x + y) &= 19 - 22 \\ y &= -3 \end{aligned}

Substitute $y=-3$ into $5x + y = 22$ :

\begin{aligned} 5x - 3 &= 22 \\ 5x &= 25 \\ x &= 5 \end{aligned}

Here both $x$ and $y$ are integers, so this value of $k$ works.

For $k=4$ :

Then $k+3=7$ and $k-1=3$ , so the system becomes

\begin{aligned} 7x + 2y &= 19 \\ 5x + 3y &= 22 \end{aligned}

Multiply the first equation by $3$ and the second by $-2$ :

\begin{aligned} 21x + 6y &= 57 \\ -10x - 6y &= -44 \end{aligned}

Add them:

\begin{aligned} (21x+6y) + (-10x-6y) &= 57 - 44 \\ 11x &= 13 \\ x &= \frac{13}{11} \end{aligned}

Here $x$ is not an integer, so $k=4$ does not work.

Conclude the correct value of k

Among the tested values $k=-1, 0, 2, 4$ , the only one that makes both $x$ and $y$ integers is $k=2$ , which gives $x=5$ and $y=-3$ .

Therefore, the correct answer is $k = 2$ (choice C).

Strategy

Hints

Desmos Guide

Step-by-step Explanation

Study Hints

Strategy

Hints

Desmos Guide

Step-by-step Explanation