Write a linear system of equations that can be used to solve these problems. Then, solve to get your final answer

1. An exam worth 145 points contains 50 questions. Some of the questions are worth two points and some are worth five points. How many two point questions are on the test? How many five point questions are on the test?.

number of two pointers --- x

number of five pointers --- y

so its says: x+y = 50
and 2x + 5y = 50

solve them please

35

x=2000

y=2

To solve this problem, we can set up a system of equations with two variables, representing the number of two-point questions and the number of five-point questions.

Let's denote:
x = number of two-point questions
y = number of five-point questions

Given that there are a total of 50 questions on the test, the first equation is:
x + y = 50

Additionally, we know that the exam is worth a total of 145 points. Since the two-point questions are worth 2 points each and the five-point questions are worth 5 points each, we have the second equation:
2x + 5y = 145

Now, we can solve this system of equations to find the values of x and y.

Method 1: Substitution Method
We can solve this system of equations by using the substitution method.

First, solve the first equation for x in terms of y:
x = 50 - y

Then substitute this expression for x into the second equation:
2(50 - y) + 5y = 145

Simplify and solve for y:
100 - 2y + 5y = 145
3y = 45
y = 15

Substitute the value of y back into the first equation to find x:
x + 15 = 50
x = 35

Therefore, there are 35 two-point questions and 15 five-point questions on the test.

Method 2: Elimination Method
Instead of solving for x in terms of y, we can eliminate one variable by multiplying the first equation by 2 and subtracting it from the second equation:

2(x + y) = 2(50)
2x + 2y = 100

Simplify the second equation:
2x + 5y = 145

Now, subtract the first equation from the second equation:
(2x + 5y) - (2x + 2y) = 145 - 100

Simplify and solve for y:
3y = 45
y = 15

Substitute the value of y back into the first equation:
x + 15 = 50
x = 35

Therefore, using the elimination method, we also find that there are 35 two-point questions and 15 five-point questions on the test.

In both methods, we have found that there are 35 two-point questions and 15 five-point questions on the test.