Chapter 4

Simultaneous Linear Equations

Section 4.1: Understanding Solutions of Simultaneous Linear Equations

In this chapter, we will study pairs of linear equations and learn how to solve them. We will discover that the solutions to two linear equations with two variables match up with the point where their graphs cross. That's because a point where two lines cross is the only point that makes both equations true at the same time.

"Simultaneous linear equations" is just a fancy name for a pair of equations that look like Ax + By = C, where A, B, and C are specific numbers (they can be positive or negative). When we say the equations are "simultaneous," we're asking: what values can we put in for x and y that make both equations true at once? Those special values are called the solutions.

For example, look at these two equations:
x + 2y = 10
x − 3y = 0

These describe two relationships between x and y. If we need both to be true at the same time, they're simultaneous equations. One solution is x = 6, y = 2, because plugging in those numbers makes both equations correct. In this chapter, we'll learn both algebra tricks and graphing methods to find these solutions.

Example 1.

Here's a super simple pair of equations: x = 21, y = 17. There's only one solution: x = 21 and y = 17. Now let's play a guessing game! Suppose these are the ages of my older brother and sister, and someone wants to figure out their ages.

I say, "The sum of their ages is 38." But that's not enough — lots of number pairs add up to 38. So I add more information: "The difference between their ages is 4."

Now the guesser can turn my clues into equations. Let's call the ages x and y. "The sum of their ages is 38" becomes:
x + y = 38

"The difference in their ages is 4" becomes:
x − y = 4

These two equations together are a pair of simultaneous linear equations. My siblings' real ages are the solution. But can the guesser actually figure out the ages from just the sum and the difference? Yes! We just need to learn how to work backwards from the sum and difference to find the original numbers.

Example 2.

Here's another example, set up like a game:

1. Pick two numbers.
2. Double one number, add the other number, and tell me the result.
3. Now swap the numbers and do the same thing again. Tell me that result too.

Someone tells me their two results: 30 and 27. After just a second, I say, "Your numbers were 8 and 11!"

How did I figure that out so fast? Let's study the pattern from Examples 1 and 2. In both cases, we start with two secret numbers. Then we do some math operations on them. In Example 1, we added the numbers together, then found their difference. So x = 21, y = 17 turned into:
x + y = 38
x − y = 4

The challenge is finding a way to work backwards. What if we add the two equations together?

(x + y) + (x − y) = 38 + 4

This simplifies to 2x = 42, so x = 21! Now let's try subtracting the two equations:

(x + y) − (x − y) = 38 − 4

This simplifies to 2y = 34, so y = 17! We found both numbers.

Let's do the same thing with Example 2. The two secret numbers are x and y. In step 2, we calculate 2x + y and get 30. In step 3, we calculate 2y + x and get 27. So now we know:

2x + y = 30
2y + x = 27

Let's add these together:

(2x + y) + (2y + x) = 57

This simplifies to 3(x + y) = 57, or x + y = 19. Now let's subtract them:

(2x + y) − (2y + x) = 30 − 27

This gives us x − y = 3. Now we know the sum and the difference of the two secret numbers, just like in Example 1! Using the same method, we find x = 11 and y = 8.

The big idea here is learning how to turn one pair of simultaneous equations into a different pair — without changing the answer. When we want to create a puzzle, we scramble the numbers using these operations. When we want to solve a puzzle, we use operations that unscramble the information. Here are our main tools:

a. Add equal amounts to equal amounts (x = 21, y = 17 became x + y = 38).

b. Subtract equal amounts from equal amounts (x = 21, y = 17 became x − y = 4).

c. Multiply both sides by the same nonzero number (like when we multiplied both sides of 2y = 34 by 1/2).

Using these three tools, we can turn one pair of equations into a different pair that has the exact same solution. It's important to remember: since we have two unknown numbers, we always need two equations to solve for them. If I only tell you that two numbers add up to 38, you can't find the exact numbers — there are too many possibilities! I'd either have to tell you one of the numbers directly, or give you another clue, like the difference between them.

The trick isn't picking operations randomly — we choose them carefully to reach our goal. In the next section, we'll see how the numbers we're given show us exactly which operations to use. There's one more tool we can use to solve equations:

d. Substitution — replacing part of one equation with an equal expression from the other equation.

Example 3.

Lovasz has 5 marbles more than twice as many marbles as Tonio has. Together, they have 107 marbles. How many marbles does Lovasz have?

Solution. First, let's use letters for our unknown numbers. Let L stand for Lovasz's marbles, and T stand for Tonio's marbles. The clue "Lovasz has 5 marbles more than twice Tonio's" turns into:

L = 5 + 2T

The second clue — that together they have 107 marbles — becomes:

L + T = 107

Since we know L equals 5 + 2T, we can substitute that into the second equation:

5 + 2T + T = 107

Now we solve like we learned in Chapter 1: combine like terms and subtract 5 from both sides to get 3T = 102, so T = 34. Tonio has 34 marbles!

To find L, we go back to the first equation and plug in T = 34:

L = 5 + 2(34), so L = 73.

We can check our work using the second equation: 73 + 34 = 107. It works!

Now let's look at how graphs can show us this same process.

Example 4.

Look at these two equations: 3x + y = 7 and x + 3y = 5. Let's graph both on the same coordinate plane and find the point where they cross.

Solution. The first line has a slope of −3, and the second has a slope of −1/3. Since the slopes are different, the lines aren't parallel — so they must cross somewhere. Using the y-intercepts — (0, 7) for the first line and (0, 5/3) for the second — we can draw both lines (see Figure 1).

The graph shows the lines crossing at the point (2, 1). Let's check this by plugging x = 2 and y = 1 into both equations:

3(2) + 1 = 7 ✓
2 + 3(1) = 5 ✓

Both equations work! In this example, we graphed the equations and read the crossing point right off the graph. Since this point lies on both lines, its coordinates (x = 2, y = 1) make both equations true. That's exactly what it means to "solve" a pair of simultaneous equations.

Example 5.

Now let's try x − 2y = 8 and 2x + 5y = 34. We'll graph both on the same grid and find where they cross.

Solution. Again, the two lines have different slopes (1/2 and −2/5), so they aren't parallel and must cross at some point. Looking at the graph (Figure 2), the lines cross at (12, 2). Since this point sits on both lines, x = 12 and y = 2 make both equations true.

Also, since two non-parallel lines can only cross at one single point, this is the only solution — there isn't another answer hiding somewhere else.

But what happens when two lines are parallel? Let's check that out next.

Example 6.

Look at these equations: 2x + 5y = 10 and 4x + 10y = 40. Let's graph them and look for a crossing point.

These two lines have different y-intercepts — (0, 2) and (0, 4) — but the exact same slope (−2/5). That means they're parallel lines that never touch! Since there's no crossing point, there are no values of x and y that make both equations true at the same time. The graph (Figure 3) confirms this — the lines never meet.

Example 7.

Now look at these equations: 2x + 5y = 20 and 4x + 10y = 40. If we graph these, we get the exact same line! That's because both equations, when rewritten in slope-intercept form, simplify to:

y = −(2/5)x + 4

In this case, there are infinitely many solutions — every single point on the line makes both equations true! Notice that one equation is just a multiple of the other (if you divide every term in the second equation by 2, you get the first equation exactly). This always happens when a pair of simultaneous equations has more than one solution.

Summary

When you're given a pair of linear equations, there are three possible outcomes:

1. Different slopes: The two lines have different rates of change (slopes), so they aren't parallel and will cross at exactly one point. That point's coordinates are the one and only solution.

1. Same slope, different intercepts: The two lines have the same rate of change but different starting points (intercepts). This means the lines are parallel and never touch — so there is no solution.

1. Same slope, same intercept: The two equations actually describe the exact same line. In this case, every point on the line is a solution — there are infinitely many answers.

In short: if two lines have different slopes, they cross at exactly one point. If two lines have the same slope, either they never cross (no solution) or they're actually the same line (many solutions).

Now let's look at what happens to the graphs when we use our operations (adding, subtracting) on simultaneous equations. Let's go back to Example 5's equations: x − 2y = 8 and 2x + 5y = 34, shown in Figure 2. If we add these two equations together, we get 3x + 3y = 42. If we subtract the first from the second, we get x + 7y = 26. When we graph these two new equations on top of Figure 2, we get Figure 4. The green and purple lines show these new equations — and you'll notice something interesting about where they cross too.