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Question 103246: Hello out there,I need some help with some problems,the first one is,Use substitution to solve:x=3y-2 and 2x+4y=16.The next one is,Use substitution to solve:2x+y=7 and 5x-2y=4.The third one is,Find the solution of the system,y=x^2+5 and y=2x+4.And the last one is,Solve the system of equations by graphing:y=x^2-4 and y=x-2.Thanks to whoever can help.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Use substitution to solve:
x = 3y - 2
2x + 4y = 16.
Substitute (3y-2) for x in the second equation, solve for y
2(3y-2) + 4y = 16
6y - 4 + 4y = 16
6y + 4y = 16 + 4
10y = 20
y = 20/10
y = 2
Find x substitute 2 for y
x = 3(2) - 2
x = 6 - 2
x = 4
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Use substitution to solve:
2x + y = 7
y = (7 - 2x); we can substitute (7-2x) for y in the next equation, find x
5x - 2y = 4
5x - 2(7-2x) = 4
5x - 14 + 4x = 4; remember a neg outside the brackets changes the signs inside
5x + 4x = 4 + 14
9x = 18
x = 18/2
x = 2
Find y using y = (7-2x)
y = 7 - 2(2)
y = 7 - 4
y = 3
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Find the solution of the system,
y = x^2 + 5
y = 2x + 4
Substitute (x^2 + 5) from the 1st equation, for y in the 2nd equation:
x^2 + 5 = 2x + 4
x^2 - 2x + 5 - 4 = 0
x^2 - 2x + 1; a quadratic equation that we can factor to:
(x-1)(x-1)
x = 1
Find y using the 2nd equation:
y = 2(1) + 4
y = 2 + 4
y = 6
Check it by substitution in the 1st equation
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Solve the system of equations by graphing:
y = x^2 - 4
y = x - 2
Plot these. substitute for x = -3 to x = +3
The first equation table
x | y
-------
-3 | +5
-2 | 0
-1 | -3
0 | -4
+1 | -3
+2 | 0
+3 | +5
Note that this is a parabola
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The 2nd equation is linear, two points are sufficient
x | y
-------
-3 |-5
+3 |+1
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The graph should look like this:
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There are two solutions, where the two graphs intersect. I'll let you figure
out what they are from the graph. When you have decided what they are, check
by substitution in the original equations.
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