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

X = # of 4 point problems Y = # 6 point problems

X + Y = 20

4X. + 6Y. = 100

Solve for where the two lines intersect for the answer.

Substitution

X + Y = 20

-X. = -X

Y. =. 20 - X

4X. + 6(20 - X) = 100

4X + 120 - 6X = 100

4X - 6X + 120 = 100

-2X + 120 = 100.

-120 = -120

-2X. =. -20

-2. -2

X. = 10

If x = 10 then plug 10 in for X in one of the problems.

X + Y = 20. 10 + Y = 20.

-10. -10

Y = 10

or

4X + 6Y = 100. 4(10) + 6Y = 100

40 + 6Y = 100

-40 -40

6Y = 60

6 6

Y = 10

Answer 2

This is 2 equations, 2 unknowns

Let x = the number of 4 point problems

Let y = the number of 6 point problems

Since there are 20 problems on the exam, the first equation is

x+y=20

Since the exam is worth 100 points, the second equation is

4x+6y=100

This system of equations can be solved using substitution or elimination. Can you solve it now that you have the equations?