Question

can someone help me with algebra1 please

Answer 1

A system of equations is given below.
2x + 7y = 1
-3x – 4y = 5

Create an equivalent system of equations by replacing the first equation by multiplying the first equation by an integer other than 1, and adding it to the second equation.
Use any method to solve the equivalent system of equations (the new first equation with the original second equation).
Prove that the solution for the equivalent system is the same as the solution for the original system of equations.

Answer 2

already did number 1. just need help with 2 and 3 please.

Answer 3

@kropot72 @experimentX @ikram002p @Hero

Answer 4

multiply the first by 3 and second by 2 and add them.

Answer 5

6x+21y=3
-6x-8y=10

Answer 6

0+13y=13

Answer 7

is that right @experimentX

Answer 8

@experimentX are you still there

Answer 9

@nikato @Hero @Sheraz12345

Answer 10

In your equivalent system, the first equation will remain the same. Only the second one will change.

Answer 11

The instructions say to multiply the 1st equation by an number other than one, then add it to the 2nd equation. It didn't say to multiply both equations by different numbers before adding them.

Answer 12

ok. so then how would i do it?

Answer 13

Follow the instructions, however, when you create your equivalent system make sure you leave the first equation as is.

Answer 14

The proper way to do it is this:

Original System:
2x + 7y = 1
-3x – 4y = 5

Multiply the first equation by 2
2(2x + 7y) = 2(1)
4x + 14y = 2

Add that result to the second equation:
4x + 14y = 2
-3x - 4y = 5

x + 10y = 7

The equivalent system becomes
2x + 7y = 1
x + 10y = 7

Answer 15

so 1 would be the answer for number 2 ?

Answer 16

You lost me. I was showing you how to create the equivalent system by following instructions.

Answer 17

oh ok. bu can you now show me the steps for number 2 ?

Answer 18

You have to solve the equivalent system, then use the solution from that to plug the values in to the original system to show that the solution to the equivalent system is also true for the original system.

Answer 19

can you please show me how? im really sleepy. i just want to get done with my homework faster

Answer 20

To solve the equivalent system:

2x + 7y = 1
x + 10y = 7

Multiply the second equation by 2:

2x + 7y = 1
2x + 20y = 14

Swapping the equations does not change the system:
2x + 20y = 14
2x + 7y = 1

Subtract the second equation from the first:
13y = 13

Solve for y:
y = 1

Solve for x:
2x + 7(1) = 1
2x + 7 = 1
2x = 1 - 7
2x = -6
x = -6/2
x = -3

So the solution to the equivalent system is (x,y) = (-3,1)

Answer 21

Now see if (-3,1) also works for the original system.

Answer 22

so (x,y) = (-3,1) is the solution for problem number 2?

Answer 23

@Hero you still there

Answer 24

When doing homework, You cannot post post a solution without showing the steps of how you arrived at the solution.

Answer 25

ok. can you show the steps with the solution

Answer 26

but where did you explain how to do part 3

Answer 27

cause i dont know how to do it @Hero

Answer 28

The solution to the equivalent system is (x,y) = (-3,1).
The original system is
2x + 7y = 1
-3x – 4y = 5

Plug the point (-3,1) in to the original system to see if it is true:
2(-3) + 7(1) = 1
-3(-3) - 4(1) = 5

-6 + 7 = 1
9 - 4 = 5

1 = 1
5 = 5

True

Answer 29

k thanks for helping