# SAT Math 4.1 Algebra and Functions

SAT Math 4.1 Algebra and Functions

 Algebra Quadratic Equations Algebra II Quadratic Equations Functions Equations and Graphs of LinearEquations and Graphs of Quadratic Language English Language Problem Solving and Data Analysis Key features of graphs Product Type SAT Math Quadratic Equations Quadratic Equations SAT Math Algebra and Functions Systems of Equations Linear Equations

### Transcript

00:22

In other words, we have to solve a system of equations.

00:26

For this problem, the easiest way to do that is by substitution because we already have

00:30

y equals something in one of the equations...

00:33

...and we can just plug in that value of y into the second equation.

00:37

We get x - 2 times the quantity -1/x squared + 4 = -2

00:44

We can distribute the -2 in the parentheses on the left side

00:48

to get x + x squared - 8 = -2.

00:52

Then we add 2 to both sides...to get x squared + x - 6 = 0.

00:59

Our equation factors into x + 3 times x - 2.

01:03

To make the left side equal zero, one of the factors has to be zero, so x equals either

01:09

2 or -3. Let's plug these two solutions into one

01:12

of our original equations to find y.

01:15

We'll go with the first one, since it's already solved for y.

01:18

When x equals 2, y equals -1/2 times 2 squared + four.

01:23

We can multiply the 2 squared by -1/2, and add 4 to get y = 2.

01:28

Next, we can plug in -3 for x, which gives us y = -1/2.

01:34

Our two solutions are (2, 2) and (-3, -1/2).

01:39

Looks like our answer is B.