SOLUTION: Your teacher is giving you a test worth 100 points containing 40 questions. There are two-point and four-point questions on the test. how many of each are on the test? K what we

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Question 104789This question is from textbook Pretice hall mathematics. Algebra 1
: Your teacher is giving you a test worth 100 points containing 40 questions. There are two-point and four-point questions on the test. how many of each are on the test?
K what we have to do is write two linear equations and solve them, I have no idea how to do it so i need help please! This question is from textbook Pretice hall mathematics. Algebra 1

Found 2 solutions by doukungfoo, Fombitz:
Answer by doukungfoo(195) About Me  (Show Source):
You can put this solution on YOUR website!
let x equal the number of two point questions and
let y equal the number of four point questions
now use the given information to write a system of equations
Given: there are 40 questions on the test
x + y = 40
Given: There are two point and four point questions for a total of 100 points
so 2 times the number of two point questions plus 4 times the number of four point questions equals 100 points
2x + 4y = 100
Now we have a system of equations we can use to solve for x and y
take the first equation and set it equal to x
x + y = 40
x = 40 - y
Now since we have shown that x equals 40 - y we can substitute that for x in the second equation
2x + 4y = 100
2(40-y) + 4y = 100
now we have an equation with one variable that we can solve for
lets do it
2(40-y) + 4y = 100
80 - 2y + 4y = 100
80 + 2y = 100
2y = 20
y = 10
Now you can use this value for y and solve for x in the first equation
x + y = 40
x + 10 = 40
x = 30
Answers: so there are 10 four point questions and 30 two point questions
Check these answers in both orginal equations:
x + y = 40
30 + 10 = 40
40 = 40
That works now try the second equation
2x + 4y = 100
2(30) + 4(10) = 100
60 + 40 = 100
100 = 100
Conclusion:
They both work so we can be sure we have correctly solved the problem.



Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Start with what you know.
Give the unknown variables a name.
T - number of two point problems
F - number of four point problems
There are 40 problems total or in algebraic terms,
1.T%2BF=40
For each two point problem, you get two points, and
for each four point problem, you get four points, and
the total value of all of the problems is 100. Or in algebraic terms,
2.2%2AT%2B4%2AF=100
Now you have two equations and two variable to solve for.
Use (1) and substitute into (2).
T%2BF=40
T%2BF-F=40-F
T=40-F
Use this result in equation (2).
2%2AT%2B4%2AF=100
2%2A%2840-F%29%2B4%2AF=100 Substitute.
80-2F%2B4F=100Distributive property.
80%2B2F=100 Simplify.
80-80%2B2F=100-80 Additive inverse of 80.
2F%2F2=20%2F2 Multplicative inverse of 2
F=10
Now use equation 1 again, remember
T=40-F
T=40-10
T=30
Check your answers.
2%2AT%2B4%2AF=100
2%2A30%2B4%2A10=100
60%2B40=100
100=100
Good answers.
30 two point problems and 10 four point problems on the test.