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Question 56982This question is from textbook
: I just don't get how to do this..here us the easiest problem on here
C=b+4c
and we have to solve for b but I just don't get what we do at all and I was even looking in the Algebra 1 cliffnotes quick review book page 50 and gives this problem as an example, solve for x and y
3x+3y=24
2x+y=13
now that lost me
it says"First multiply the bottom equation by 3. Now the y is preceding by a 3 in each equation"
so if you you multiply by 3 then how do you get 2x+y=13? Can somebody please tell me how to do this, my teacher doesn't make any since at all.
This question is from textbook
Answer by Earlsdon(6294)   (Show Source):
You can put this solution on YOUR website! 1. Solve C = b + 4c for b
To solve for b means to get the b on one side of the equation and everything else on the other side.
C = b + 4c Subtract 4c from both sides of the equation.
C - 4c = b + 4c - 4c Simplify.
C - 4c = b or b = C - 4c
2. Solve for x and y:
1) 3x + 3y = 24
2) 2x + y = 13
This is called a system of two equations in two unknowns (variables). There are several methods available for solving these but it appears that "elimination" is the method suggested by your book. By multiplying the second equation by 3, you do get the y precedes by 3 in both equations as seen below:
3(2x + y = 13) Applying the distributive property, you get equation 2a):
2a) 6x + 3y = 39 Now you subtract equation 1) from equation 2a) to "eliminate" the y's thus leaving you with one equation in one unknown (variable).
2a) 6x + 3y = 39
1) -(3x + 3y = 24) and you get:
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3x = 15 Now divide both sides by 3.
x = 5 So now you have the value of x, you can substitute this into either of the original equations and solve for y. Let's use equation 2)
2) 2x + y = 13 Substitute x = 5
2(5) + y = 13 Simplify.
10 + y = 13 Subtract 10 from both sides.
y = 3
The answer is:
x = 5 and y = 3 or (5, 3)
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