SAT Math 7.3 Algebra and Functions

SAT Math 7.3 Algebra and Functions

Algebra and FunctionsOperations on Algebraic Expressions
Basic AlgebraAlgebraic Expressions
Heart of AlgebraSystems of linear equations in two variables (word problems)
LanguageEnglish Language
Pre-AlgebraBasic Operations
Product TypeSAT Math
SAT MathAlgebra and Functions

Transcript

00:21

And here are the potential answers...

00:26

See those t's in the x and y equations?

00:28

This shows us that we're dealing with something

00:31

called PARAMETRIC equations.

00:33

It's basically where we write our x and y equations in terms of another variable...like t.

00:41

This question asks us to solve for y...if we look at our answer choices, they want us

00:45

to solve for y in terms of x.

00:48

Well, we can do this by solving for t in terms of x and then substituting

00:51

the t value in terms of x back into the y equation.

00:55

Rearranging the first equation so that  t is by itself on one side...

01:00

we get t equals x - 1.

01:02

We can just insert x - 1 into the equation now wherever we see t...

01:08

x - 1 squared is the same as x - 1 times x - 1...

01:14

...which equals x squared - x - x + 1.

01:20

When we add like terms, we get x squared - 2x + 1.

01:27

That value times 3...gives us 3x squared - 6x + 3.

01:31

2 times the quantity x - 1 equals 2x - 2...

01:37

Combining like terms...we have 3x squared - 6x + 2x

01:42

and that gives us - 4x...3 - 2 - 1...gives us 0.

01:49

So we're left with 3x squared - 4x.

01:51

Which is answer choice (C).