1. 7.1 Linear and Nonlinear Systems of Equations
Substitution
Solving a System of 2 equations in 2 variables.
x + y = 4
x - y = 2
Solve for x or y in the easiest way possible and in either equation.
From eq y = 4 - x
Substitute 4 - x in for y in equation 2.
x - (4 - x) = 2
x x = 2
2x = 6
x = and by backsubstituting, y = 1

2. Note: multiply this equation by 100 so we don’t have any decimals.
A total of $12,000 is invested in two funds paying 9% and 11% simple interest. If the yearly interest is $1,180, how much of the $12,000 is invested at each rate?
So what are the two equations?
x + y = 12, two funds totaling $12,000
.09x + .11y = 1, Interest formula. RP = I
Note: multiply this equation by 100 so we don’t have any decimals.
Solving for x in equation 1 we have: x = 12,000 - y
Substituting, we have
9(12,000 - y) + 11y = 118,000
108, y + 11y = 118,000
2y = 10, and y = $5,000 and x = $7,000

3. Solve: x2 - x - y = 1
-x + y = -1
Solve for y in equation 2
y = x - 1
Now, substitute into eq. 1
x2 - x - (x - 1) = 1
x2 - 2x + 1 = 1
x2 - 2x = Factor x(x - 2) = 0
x = 0, x = 2
Now, we need to back substitute to find our y values.
For x = 0, y = -1
For x = 2, y = 1
Note: We must check all answers to
make sure that they all work. These do.

4. A small business invests $10,000 in equipment to produce a product
A small business invests $10,000 in equipment to produce a product. Each unit of the product costs $0.65 to produce and is sold for $ How many items must be sold before the business breaks even?
The total cost of producing x units is what?
C = 0.65x + 10,000
The revenue obtained by selling x units is what?
R = 1.2x
The break-even point occurs when R = C. Hence,
1.2x = 0.65x + 10,000
.55x = 10, and x = 18,182 units