SAT Math 4.1 Algebra and Functions
|Algebra II||Quadratic Equations|
|Functions||Equations and Graphs of Linear|
Equations and Graphs of Quadratic
|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.