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more algebra 2 help :b

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MZ132613

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*different algebra problems :b i still need some help though
can pay with EOC RSGP if the answers are right :)

Question 24
Solve y = 2x – 4
7x – 5y = 14
by the substitution method.
A)(2, 0)

B)(3, 2)

C)(-2, 0)

D)(0, 2)


Question 25
Solve the following system of equations by using the substitution method.
x + y = 1
3x – y = 11

A)(-3, 4)

B)(6, -5)

C)(3, -2)

D)Inconsistent


Question 26
Solve the following system of equations by using the substitution method.
x – y = 3
2x + 2y = 2

A)(3, 4)

B)(-4, 5)

C)(2, -1)

D)Inconsistent


Question 27
What would be the first step to solve this system of equations by the Substitution Method?
3x + 4y = -1
x + 3y = 3

A)Solve the second equation for x.

B)Divide the first equation by 4.

C)Add the two equations together.

D)Multiply the second equation by -3.


Question 28
What would be the first step to solve this system of equations by the Substitution Method?
10x – 2y = 8
4x – 10y = -6

A)Add the two equations together.

B)Solve the first equation for y.

C)Multiply the first equation by -5

D)Multiply the second equation by 5


Question 29
What would be the first step to solve this system of equations by the Substitution Method?
2x + y = 7
5x – 2y = -5

A)Solve the first equation for y.

B)Multiply the first equation by 2.

C)Multiply the second equation by 5.

D)Add the two equations together.


Question 30
Solve
2x + 3y = 18
5x – y = 11
by the elimination method.

A)(0, 6)

B)(3, 4)

C)(9, 0)

D)(-3, -4)


Question 31
Solve the following system of equations by using the elimination method.
x – 4y = 17
3x + 4y = 3

A) (7, -5/2)

B)(7/2, -27,4)

C)(5, -3)

D)(7, -6)


Question 32
Solve
9x + 2y = 2
21x + 6y = 4
by the elimination methods.

A)(1/3, -1/2)

B)(3, -2)

C)(-3, 2)

D)(-1/3, 1/2)


Question 33
Solve the following system of equations by using the elimination method.
2x + 6y = 4
3x – 2y = -16

A)(-4, 2)

B)(2, -4)

C)(-12/11, 68/11)

D)(52/11, -60/11)


Question 34
What would be the first step to solve this system of equations by the Elimination Method?
3x + y = 16
2x + 5y = 15

A)Multiply the second equation by 5.

B)Solve the first equation for y.

C)Multiply the first equation by -5.

D)Add the two equations together.


Question 35
What would be the first step to solve this system of equations by the Elimination Method?
3x + 4y = -1
x + 3y = 3

A)Solve the second equation for x.

B)Divide the first equation by 4.

C)Add the two equations together.

D)Multiply the second equation by -3.


Question 36
What would be the first step to solve this system of equations by the Elimination Method?
2x – 3y = 22
4x + 5y = -22

A)Add 3y to both sides of the first equation.

B)Multiply the second equation by 2.

C)Multiply the first equation by -2.

D)Divide the first equation by 5.


*You don't have to show work, but if you do that'd be really nice just so i know how you did it and stuff, but it's optional :b

you can also add me on Skype if that will help: MZ132613
 
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The reason you need substitution/elimination for these questions is because you can't find the values of x or y if both variables are present in the equation (i.e. in x+y = 1 there are infinite combinations of two numbers that equal 1 so you can't solve for x and y).

Substitution/elimination lets you change the equation to have 1 variable instead of 2 to make it simple to solve.

So in the first example with substitution:
y = 2x – 4
7x – 5y = 14

If you wanted to solve for x using the second equation you'd need to re-write its the equation so that it only contains constant terms (1,2,3 etc.) and terms containing the x variable so that it's nice and easy to solve for x. From the first equation you know that y is equal to 2x-4 meaning that 7x-5(2x-4) is the same as 7x-5y.

You can then rewrite the second equation as
7x-5(2x-4) = 14
7x-10x+20=14
-3x = -6
x = 2

Since you know x is equal to 2 you can rewrite either equation with x equalling to 2.
y = 2(2)-4
y = 0

Or

7(2)-5y = 14
-5y = 14 - 14
-5y = 0
y = 0

Hope that helped you get a better understanding of how to approach these questions.
 
win4fun said:
The reason you need substitution/elimination for these questions is because you can't find the values of x or y if both variables are present in the equation (i.e. in x+y = 1 there are infinite combinations of two numbers that equal 1 so you can't solve for x and y).

Substitution/elimination lets you change the equation to have 1 variable instead of 2 to make it simple to solve.

So in the first example with substitution:
y = 2x – 4
7x – 5y = 14

If you wanted to solve for x using the second equation you'd need to re-write its the equation so that it only contains constant terms (1,2,3 etc.) and terms containing the x variable so that it's nice and easy to solve for x. From the first equation you know that y is equal to 2x-4 meaning that 7x-5(2x-4) is the same as 7x-5y.

You can then rewrite the second equation as
7x-5(2x-4) = 14
7x-10x+20=14
-3x = -6
x = 2

Since you know x is equal to 2 you can rewrite either equation with x equalling to 2.
y = 2(2)-4
y = 0

Or

7(2)-5y = 14
-5y = 14 - 14
-5y = 0
y = 0

Hope that helped you get a better understanding of how to approach these questions.
Click to expand...
oh i gotcha now. That helped a lot, thank you, I understand it a lot better :D
 
Using the substitution method, just look at it like this. (X,Y) where you put the number in the first slot (x) substituted as "x" so for your first question's answer. You would find (A) correct.

What it looks like when substituted: [y = 2x - 4] being 0 = 2(2) - 4 [Which is 2 times 2 minus 4, which is 0]
and [7x - 5y = 14] being 7(2) - 5(0) = 14 [Which is 7 times 2 minus 5 times 0, which is 14]

Hope that simplifies the simplicity of these substitution problems :)
 
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