Question 103246
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:
{{{ graph( 300, 200, -6, 6, -6, 6, x^2-4, x-2) }}}
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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.