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Teachers & SchoolsSAT Math 5.5 Algebra and Functions

Algebra | Factoring |

Algebra and Functions | Solving Quadratic Equations by Factoring |

Functions | Build new functions from existing functions Quadratic Functions |

Language | English Language |

Math | Algebra |

Passport to Advanced Math | Solving quadratic equations |

Product Type | SAT Math |

SAT Math | Algebra and Functions |

Well, the first function we’re given, x squared minus 100 over x squared plus 10x, can be factored.

X squared minus 100 looks a heck of a lot like the difference of two squares….

so the numerator becomes (x + 10) times (x - 10). And the denominator is x squared plus 10x…

which becomes x times (x + 10). There’s an x + 10 now in both the numerator

and denominator, so we can cancel out the x +10 in each place.

Our new equation looks like this: f of x equals x minus 10 over x.

We’re given that f of x minus 5 equals 1/3, so substituting x minus 5 into the first function,

we get f of x minus 5… equals x minus 5 minus 10…all over x minus 5… equals 1/3.

X minus 5 minus 10 simplifies to x minus 15. So x minus 15 over x minus 5 equals 1 over 3.

To eliminate the fractions on both sides,

we need to multiply both sides by (x - 5) and 3… to arrive at 3 times (x - 15) = x - 5.

We distribute the 3 so that 3x - 45 = x – 5…and solve.

And x equals 20. Our answer’s E.

And our messy functions are no more. Thank you, Mr. Clean.