High School Algebra II Unlocked (2016)
Chapter 2. Systems of Equations and Rational Expressions
Lesson 2.2. Systems of Equations
REVIEW
A system of equations consists of two or more equations. Any solution to a system of equations is a solution to each of the individual equations in the system.
One more way to solve a single-variable polynomial equation is by creating and solving a system of equations in two variables.
A solution to a system of equations is a coordinate pair (x, y) that makes both equations true. Each equation in the system shows y equal to a polynomial in terms of x, so the x-value of a solution to the system represents a solution to the equation where the two polynomials of x terms are set equal to each other.
Solve the equation 1/2 x5 + x4 − 3/2 x3 − x2 = 5x + 6.
We can rewrite this as a system of equations by setting each side of the equation equal to y.
y = 1/2 x5 + x4 − 3/2 x3 − x2
y = 5x + 6
An x-value that produces the same y-value in each equation makes 1/2 x5 + x4 − 3/2 x3 − x2 equal to 5x + 6. In other words, it will be a solution to our original single-variable equation. Using graphing technology, we can graph y = 1/2 x5 + x4 − 3/2 x3 − x2 and y= 5x + 6, as shown below.
Alternatively, you could
rewrite the equation in
any arrangement you
like, such as 1/2 x5 + x4 =
3/2 x3 + x2 + 5x + 6. This
would produce a different
system of equations,
but the solutions for x
would still be the same.
A graphing calculator
allows you to be
precise about points of
intersection. However,
when looking at a graph
with unlabeled points, you
cannot assume that the
function graphs intersect
exactly at (−3, −9). It could
be (−2.9, −8.9), which
is why it’s important to
check your solutions.
The points where the graphs intersect are the solutions to the system of equations: (−3, −9), (−1, 1), and (2, 16). Let’s confirm these solutions using the equations.
When x = −3, 1/2 (−3)5 + (−3)4 − 3/2 (−3)3 − (−3)2 = −243/2 + 81 + 81/2 − 9 = −9 and 5(−3) + 6 = −15 + 6 = −9.
When x = −1, 1/2 (−1)5 + (−1)4 − 3/2 (−1)3 − (−1)2 = −1/2 + 1 + 3/2 − 1 = 1 and 5(−1) + 6 = −5 + 6 = 1.
When x = 2, 1/2 (25) + 24 − 3/2 (23) − 22 = 16 + 16 − 12 − 4 = 16 and 5(2) + 6 = 10 + 6 = 16.
The points (−3, −9), (−1, 1), and (2, 16) are indeed solutions to the system of equations, so their x-values (−3, −1, and 2) are solutions to the equation 1/2 x5 + x4 − 3/2 x3 − x2 = 5x + 6.
Since this polynomial equation is of degree 5, it has a total of five solutions, but we have only found three so far. Let’s move all terms to one side and solve for the remaining zeros.
1/2 x5 + x4 − 3/2 x3 − x2 − 5x − 6 = 0
Our solutions so far correspond to factors (x + 3), (x + 1), and (x − 2), which multiply together to produce x3 + 2x2 − 5x − 6. Let’s divide 1/2 x5 + x4 − 3/2 x3 − x2 − 5x − 6 by x3 + 2x2 − 5x − 6 to see what factors remain.
See Example 18 in
Chapter 1, Lesson 1.5,
for a walk-through of
how to perform long
division of polynomials.
The other factor is (1/2 x2 + 1). The roots of this quadratic are ±i 
.
The five solutions to 1/2 x5 + x4 − 3/2 x3 − x2 = 5x + 6 are x = −3, x = −1, x = 2, x = −i 
, and x = i 
.
Here is how you may see systems of equations on the SAT.
y = −2x + 3
y = 2x2 − 12x + 11
How many values of (x, y) satisfy the system of equations above?
A) 0
B) 1
C) 2
D) Infinitely many