ACT Math Systems of Linear Equations

ACT Math: Systems of Linear Equations

A system of equations is a set of two or more equations that have two or more variables. “Solving” the system of equations means finding values for each of the variables that make each equation true. You probably remember from school that there are a few different ways to solve a system of equations. This shouldn’t deter you. While it takes some practice to choose the quickest method for solving a particular system of equations, you’ll eventually find that having options greatly increases your speed on the test. Let’s check out some methods.

Addition or Subtraction

In some rare cases, you can simply add or subtract both equations to isolate a variable. If the equations are set up the right way, addition/subtraction is by far the quickest choice.

Solve:           x+y=10

x –y= 2

Let’s add:   2x= 12→ x=6

Plug in 6 for x: 6+y=10→ y=4

Notice how fast that was. Because the y’s canceled upon addition, we quickly isolated the x variables in order to solve for x. If one of the y’s had a coefficient of, say, 7, then we could have chosen to subtract the equations and thus cancel the x’s.

Substitution

Substitution might be the method you’re most comfortable with. It’s called ‘substitution’ because you are substituting a variable in one equation (say, x) for an expression derived from the other equation (say, y+5); in this way, you write one equation in terms of only one variable, rendering it solvable.

Solve: 2x+ y = 10

-6x + 2y = 12

Let’s solve for y in the first equation:2x+y=10→ y=10-2x

Now, let’s substitute 10-2x for y in the other equation: -6x+2(10-2x)=12→ -6x+20-4x=12

→ -10x+20=12è10x=8 → x=8/10→ x=4/5

Now, let’s plug in 4/5 for x in the first equation (because it’s simpler): 2(4/5)+y=10→ 8/5+y=10

→ y=10- (8/5)→ y=50/5  –  8/5 → y=42/5

Elimination

Elimination is essentially solving by addition/subtraction, but it requires a little more manipulation. Basically, elimination is manipulating a system of equations so that the addition or subtraction will become a viable method.

Solve:    x+2y = 10

4x+y  =  8

Basically, we can multiply either of the equations by some constant in order to make the variables cancel upon addition or subtraction. In this case, we can either multiply the top equation by -4 (to cancel the x’s) or multiply the second equation by -2 (to cancel the y’s); let’s go for the second option.

X+2y=10             x+2y=10

-2(4x+y=8)→     -8x-2y=-16

Now, add: -7x=-6→ x=6/7

Lastly, substitute 6/7 for x in one of the original equations. I’ll choose the first:

6/7   + 2y=10 → 2y=70/7  –  6/7 → 2y=  64/7 →  y=64/7 *  ½  → y=64/14 → y=32/7

Tricky Problems

Some systems of equations problems will not ask you to solve, but to perform some other calculation. Let’s see an example:

If 3a+5b=10 and 5a+3b=30, what is the average of a and b?

First, set them up:  3a+5b=10

5a+ 3b=30

Next, consider what average means. The average of two variables is their sum (a+b) divided by 2 (the number of items). So, what we really want to solve for is a+b. We could get at this by using elimination and substitution, but look how fast we can solve using addition.

3a+5b = 10

+ 5a + 3b = 30

8a + 8b = 40

Next, divide by 8 to yield a+b:

a+b = 5

If a+b is 5, then the average of a and b is 5/2.