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Teachers & SchoolsSAT 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) |

Language | English Language |

Pre-Algebra | Basic Operations |

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).