Question 268·Hard·Algebra

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A system of linear equations is given by

\begin{cases} 2x + 5y = 19 \\ kx - y = -1 \end{cases}

If $x$ and $y$ are both integers and $k$ is an integer with $1 \le k \le 10$ , what is the value of $k$ ?

Your answer

An integer solution to a system with an unknown integer coefficient needs more than a point where the lines cross. Isolate one variable and substitute to get an integer times an expression containing the coefficient. That expression must divide the number on the other side. Use the coefficient’s allowed range to narrow the divisors, then check that both coordinates are integers.

Hints

Hint 1
The second equation has $-y$ , so substitution is a good start: get $y$ alone and put the entire expression in place of $y$ in the first equation. What will multiply $x$ ?
Hint 2
After collecting the $x$ -terms, you’ll have an integer times integer $x$ equal to a fixed number. A divisor goes into that number evenly. What must be true of the expression multiplying $x$ ?
Hint 3
Use both restrictions on $k$ . Its range limits the possible divisors, and integer $k$ makes $5k+2$ equal to $2$ more than a multiple of $5$ . Which divisor passes both tests?

Step-by-step

Substitute, then use integer divisibility
Step 1
Get $y$ alone
Use substitution: replace $y$ in one equation with an equal expression from the other. Start with $kx-y=-1$ . Add $y$ to both sides: $kx=y-1$ . Add $1$ to both sides: $kx+1=y$ . Now the whole expression $kx+1$ can replace $y$ .
Step 2
Substitute into the first equation
Replace $y$ with all of $kx+1$ in $2x+5y=19$ : $2x+5(kx+1)=19$ . Keep the parentheses because the $5$ multiplies both terms inside them.
Step 3
Find the number multiplying $x$
Distribute $5$ : $2x+5kx+5=19$ . Collect the $x$ -terms into one coefficient of $x$ : $(5k+2)x+5=19$ . Subtract $5$ : $(5k+2)x=14$ . This form lets you use the condition that $x$ is an integer.
Step 4
Narrow the possible divisors
Since $x$ is an integer, $5k+2$ must divide $14$ evenly. A divisor is a whole number that goes into another whole number with no remainder. Use $1\le k\le10$ to bound $5k+2$ : $7\le5k+2\le52$ . The positive divisors of $14$ are $1,2,7,14$ , so only $7$ and $14$ fit that range. Because $k$ is an integer, $5k+2$ must be $2$ more than a multiple of $5$ . Check the two candidates: $7-2=5,\qquad14-2=12$ . Only $5$ is a multiple of $5$ , so:
$5k+2=7.$
Step 5
Find $k$ and check both coordinates
Type $5k+2\sim7$ in Desmos. The $\sim$ asks it to find $k$ rather than graph an equation; under PARAMETERS, Desmos shows $k=1$ . Then type both original equations and click their intersection. Desmos shows $(2,3)$ , so $x$ and $y$ are both integers. The value of $k$ meets every condition. Grid in 1.

Question 268·Hard·Algebra

Strategy

Hints

Step-by-step

Lessons that teach this

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Strategy

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Step-by-step

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