Your next algebra has a mix 20 problems. Some problems are worth 4 points and others are worth 6 points. The test is worth 100 points total. How many of each type of problem are in the test?

Answer 1

Let x=problems worth 4 points. Let y=problems worth 6 points. We know there are 20 total problems, so x+y=20. We also know there are a total of 100 points. So, 4x+6y=100. Now we solve the system of equations. Multiply (x+y=20) by -4. That equation is now -4x-4y= -80. Add the system of equations. 4x + (-4x) = 0. 6y+(-4y) = 2y. 100 + (-80) = 20. So 2y=20, y=10. Now go back to one of the original equations, x+y=20. We know y=10, so we have x+10=20, so x=10. So there are 10 of each type of problem.