SAT Math : How to find the equation of a curve
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Example Question #4 : How To Find The Slope Of Perpendicular Lines
Solve the equation for x and y.
x² + y = 31
x + y = 11
x = 8, –6
y = 13, 7
x = 6, 15
y = 5, –4
x = 13, 7
y = 8, –6
x = 5, –4
y = 6, 15
x = 5, –4
y = 6, 15
Solving the equation follows the same system as the first problem. However since x is squared in this problem we will have two possible solutions for each unknown. Again substitute y=11-x and solve from there. Hence, x2+11-x=31. So x2-x=20. 5 squared is 25, minus 5 is 20. Now we know 5 is one of our solutions. Then we must solve for the second solution which is -4. -4 squared is 16 and 16 –(-4) is 20. The last step is to solve for y for the two possible solutions of x. We get 15 and 6. The graph below illustrates to solutions.
Example Question #5 : How To Find The Slope Of Perpendicular Lines
Solve the equation for x and y.
x² – y = 96
x + y = 14
x = 25, 4
y = 10, –11
x = 15, 8
y = 5, –14
x = 5, –14
y = 15, 8
x = 10, –11
y = 25, 4
x = 10, –11
y = 25, 4
This problem is very similar to number 2. Derive y=14-x and solve from there. The graph below illustrates the solution.
Example Question #1 : X And Y Intercept
Solve the equation for x and y.
5x² + y = 20
x² + 2y = 10
x = √10/3, –√10/3
y = 10/3
No solution
x = 14, 5
y = 4, 6
x = √4/5, 7
y = √3/10, 4
x = √10/3, –√10/3
y = 10/3
The problem involves the same method used for the rest of the practice set. However since the x is squared we will have multiple solutions. Solve this one in the same way as number 2. However be careful to notice that the y value is the same for both x values. The graph below illustrates the solution.
Example Question #3 : How To Find The Slope Of Perpendicular Lines
Solve the equation for x and y.
x² + y = 60
x – y = 50
x = 10, –11
y = –40, –61
x = 11, –10
y = 40, 61
x = 40, 61
y = 11, –10
x = –40, –61
y = 10, –11
x = 10, –11
y = –40, –61
This is a system of equations problem with an x squared, to be solved just like the rest of the problem set. Two solutions are required due to the x2. The graph below illustrates those solutions.
Example Question #1 : X And Y Intercept
A line passes through the points 
None of the available answers
First we will calculate the slope as follows:
And our equation for a line is
Now we need to calculate b. We can pick either of the points given and solve for
Our equation for the line becomes
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