﻿ ﻿SOLVING LINEAR-QUADRATIC SYSTEMS - Equations and Inequalities - ALGEBRA - SAT SUBJECT TEST MATH LEVEL 1

## CHAPTER 6Equations and Inequalities

A question on the Math 1 test could ask you to solve a system of equations in which one, or even both, of the equations are quadratic. The next example illustrates this.

EXAMPLE 30: To solve the system, use the substitution method. Replace the y in the second equation by 2x – 1.

If x = 3, then y = 2(3) – 1 = 5; and if x = 1, then y = 2(1) – 1 = 1.

So there are two solutions: x = 3, y = 5 and x = 1, y = 1.

As you will see in Chapter 13, solving the system of equations in Example 30 is equivalent to determining the points of intersection of the line y = 2x – 1 and the parabola y = x 2 – 2x + 2. Those points are (1, 1) and (3, 5).

It follows from the discussion in the previous paragraph that an alternative method of solving the system of equations in Example 30 is to graph them. If you have a graphing calculator, you can graph the given line and parabola and then determine their points of intersection. Which solution is preferable? This is a personal decision. If your algebra skills are strong, solving the system graphically is no faster and offers no advantage. If, on the other hand, your algebra skills are weak and your facility with the calculator is good, you should avoid the algebra and use your calculator.

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