Question 137·Medium·Systems of Two Linear Equations in Two Variables
The system of equations is
The solution to the system is . What is the value of ?
For systems where the coefficients of one variable are opposites (like and ), use elimination: add the equations to cancel that variable and solve for the remaining one. Then substitute back into either original equation, but focus on the quantity the problem actually asks for (here ), so you can stop as soon as that value is found instead of doing extra work to solve for both and if it’s not necessary.
Hints
Look at how the -terms line up
Compare the -terms: one equation has and the other has . What happens to if you add the two equations together?
Solve for one variable first
After you eliminate by adding the equations, you will get a simple equation in terms of . Solve that equation to find .
Substitute to get
Once you know , plug it into either original equation and simplify carefully. Pay attention: you only need the value of , not .
Desmos Guide
Enter the system of equations
In Desmos, type the two equations exactly as they are, each on its own line:
5x - 2y = 11-5x + 7y = 9Desmos will graph both lines.
Find the intersection point
Click on the point where the two lines intersect. Desmos will display the coordinates of this intersection, which is the solution to the system.
Compute the requested value
In a new expression line, type 5 * ( followed by the -coordinate of the intersection, then ). The output is the value of you need to choose from the answer options.
Step-by-step Explanation
Use elimination to remove one variable
Notice the in the first equation and the in the second:
If you add the two equations, the and terms cancel out. This is the elimination method.
Add the equations and solve for
Add the left sides and the right sides:
Combine like terms:
So you get
Now solve for :
Substitute to find an equation for
Substitute into one of the original equations, for example :
Compute :
Now isolate by adding 8 to both sides:
Answer the question
The problem asks for the value of , not itself. From the previous step, evaluate , so . The correct choice is 19.