Systems of Equations Part 2: How Can I Use This Skill?

In practice, systems of equations are used mostly in STEM-related fields. They are part of such tasks as building computer models (like weather and climate forecasting models) and simulating complex phenomena. 

For example, here are the equations that model how heat is transferred by convection. Convection is a method of heat transfer in a fluid through movement/circulation of the warmer parts of the fluid. Hot air rises and cold air sinks because hot air is less dense than cold air. If the bottom part of a fluid is heated (like a pot of water on a hot stove), then the hotter part of the fluid is below the cooler part. The hotter fluid (in this case water) will then rise to the top while the cooler water will sink to the bottom. Because of the heat source, the cooler water will now be heated up while the warmer water will cool because it is further away from the stove. The cycle then repeats itself.

An illustration of how convection works in liquids.

Image by Zachary Wilson

As you can see, systems of equations get much more complicated than the ones you’ll be solving on the ACT®. The skills in this unit are fundamental ones that you can build on.

Systems of Equations Part 3: Guided Practice

Colored circles on a color-blocked background.

Image by Sarah Pflug

1. Solve the following system through substitution.

You have a pile of red, pink, and yellow Starbursts. You have 5 more red Starbursts than pink Starbursts. You have 3 times as many pink Starbursts as yellow Starbursts. If you have a total of 26 Starbursts, how many yellow ones do you have?

Let’s write equations for each type of starburst, then substitute them into one, longer equation.

P = pink Starbursts, R = red Starbursts, and Y = yellow Starbursts

If you have 5 more red Starbursts than pink Starbursts, P + 5 = R.

If you have 3 times as many pink Starbursts as yellow Starbursts, \( Y= \frac{P}{3} \).

If your total number of Starbursts can be written as P + R + Y = 26, then we can substitute P + 5 in for R and \(\frac{P}{3} \) in for Y.

\( P + P + 5 + \frac{P}{3} = 26 \)

If we multiply the equation by 3, we get:

3P + 3P + 15 + P = 78.

7P = 63

P = 9

If you have 9 pink Starbursts, then you must have \(\frac{9}{3}\) = 3 yellow Starbursts.

2. Solve the following system through graphing and elimination.

8y – 5x = 39

4y – 2x = 19

Graphing:

A graph of 8y - 5x = 39 and 4y - 2x = 19

The lines intersect at (-1, 4.25), so x = -1 and y = 4.25. 

Elimination:

8y – 5x = 39

4y – 2x = 19

First, let’s turn the 4y in the second equation into a -8 y by multiplying the entire equation by -2. The second equation is now -8y + 4x = -38. Next, we will add the two equations together.

   8y – 5x = 39

-8y + 4x = -38

           -x = 1

x = -1. We can plug -1 in for x in either of the original equations to find that y = 4.25.

3. Solve the following system by setting equal.

16x + 19 = y

-4x + 14 = y

Set the two equations equal to each other.

16x + 19 = -4x +14

Combine like terms.

20x = -5

Solve for x.

x = \( -\frac{1}{4} \)

Plug this value for x into either equation to solve for y.

\( 16 * -\frac{1}{4} + 19 = -4 + 19 = 15 \).

y = 15.

4. How many real solutions does the following system of equations have? 

x + 3y = -9

\( x^2 – 6 + 4y = -13 \)

Find the answer through graphing and substitution.

Graphing:

A graph of a system of equations with two solutions.

We can see from the graph of the functions that they cross two times, so there are two

real solutions.

Substitution:

x + 3y = -9

\( x^2 – 6 + 4y = -13 \)

You can solve the first equation for either x or y, then substitute into the second equation. You will get the same answers either way.

Solve the first equation for x.

x = -9 – 3y

Substitute this value for x in the second equation.

\( (-9 – 3y)^2 -6 +4y = -13 \)

FOIL and combine like terms:

81 + 54y + 9y^2 – 6 + 4y = -13

9y^2 +58y + 88 = 0

Plug this into the quadratic formula.

\( \frac{-58 \pm \sqrt{58^2 – 4*9*88}}{2*9} = 0 \)

The part of the quadratic formula under the square root (\( b^2 – 4ac \)), is called the discriminant. If this number is positive when simplified, a quadratic has 2 real roots. If it is 0, the quadratic has 1 real root. If it is negative, the quadratic has no real roots.

Look at the discriminant and simplify.

\( 58^2 – 4*9*88 = 3364 – 3168 = 196 \)

Because 196 is positive, there are 2 real solutions

 

If you think you’ve got the hang of it, try our 5-question quiz!