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Teachers & SchoolsSAT Math 4.1 Algebra and Functions

Algebra | Quadratic Equations |

Algebra II | Quadratic Equations |

Functions | Equations and Graphs of Linear Equations 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 |

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

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

y equals something in one of the equations...

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

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

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

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

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

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

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

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

of our original equations to find y.

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

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

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

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

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

Looks like our answer is B.