SAT Math 7.3 Algebra and Functions
|Algebra and Functions||Operations on Algebraic Expressions|
|Basic Algebra||Algebraic Expressions|
|Heart of Algebra||Systems of linear equations in two variables (word problems)|
|Product Type||SAT Math|
|SAT Math||Algebra and Functions|
And here are the potential answers...
See those t's in the x and y equations?
This shows us that we're dealing with something
called PARAMETRIC equations.
It's basically where we write our x and y equations in terms of another variable...like t.
This question asks us to solve for y...if we look at our answer choices, they want us
to solve for y in terms of x.
Well, we can do this by solving for t in terms of x and then substituting
the t value in terms of x back into the y equation.
Rearranging the first equation so that t is by itself on one side...
we get t equals x - 1.
We can just insert x - 1 into the equation now wherever we see t...
x - 1 squared is the same as x - 1 times x - 1...
...which equals x squared - x - x + 1.
When we add like terms, we get x squared - 2x + 1.
That value times 3...gives us 3x squared - 6x + 3.
2 times the quantity x - 1 equals 2x - 2...
Combining like terms...we have 3x squared - 6x + 2x
and that gives us - 4x...3 - 2 - 1...gives us 0.
So we're left with 3x squared - 4x.
Which is answer choice (C).