Chapter 4

Simultaneous Linear Equations

Section 4.1: Understanding Solutions of Simultaneous Linear Equations

Analyze and solve pairs of simultaneous linear equations. Understand that the solutions to a system of two linear equations in two variables correspond to the points where their graphs cross, because a point of intersection satisfies both equations at once. (Standard 8.EE.8a)

Simultaneous linear equations are a pair of equations, each in the form Ax + By = C, where A, B, and C are specific numbers (positive or negative). Calling them "simultaneous" raises a question: are there values for the variables (x and y) that make both equations true at the same time? Any pair of values that does this is called a solution of the simultaneous equations.

For example, x + 2y = 10 and x − 3y = 0 describe two separate relationships between x and y. If we require both to be true together, they're simultaneous equations. One solution is x = 6, y = 2, since plugging these values in makes both equations true. In this chapter, we'll explore both algebraic and graphical methods for finding solutions to simultaneous linear equations.

Example 1.

x = 21, y = 17 is technically a pair of simultaneous linear equations, and clearly it has just one solution: x = 21, y = 17. Now suppose I want to disguise this fact. Say these numbers are the ages of my older siblings, and you ask me for their ages. Instead of answering directly, I say: "The sum of their ages is 38." That's not enough — many pairs of numbers add up to 38. So I add another clue: "The difference in their ages is 4." Now can you figure out their ages?

If you call the ages x and y, you can turn my two clues into equations. "The sum of their ages is 38" becomes x + y = 38. "The difference in their ages is 4" becomes x − y = 4. Together:

x + y = 38
x − y = 4

These two equations are a pair of simultaneous linear equations, and my siblings' actual ages are a solution. But is this enough information to find that solution? If we know the sum and the difference of two numbers, can we always find the numbers themselves? The answer is yes — and solving this problem means figuring out how to work backward from the sum and difference to the original numbers.

Example 2.

Here's another example, presented as a game, that shows the same idea:

1. Pick two numbers.
2. Double one number and add the other. Tell me the result.
3. Now switch the numbers and repeat step 2. Tell me that result too.

Suppose you tell me the results are 30 and 27. After a moment, I announce: "Your two numbers were 8 and 11."

How did I solve it so quickly? Let's break down what's happening. In both Examples 1 and 2, we start with a specific pair of numbers, then perform algebraic operations on them. In Example 1, we added the numbers together and then subtracted them: from x = 21, y = 17, we produced x + y = 38 and x − y = 4. The challenge is to reverse this process.

One approach: add the two equations together.

x + y = 38
x − y = 4

Adding gives us:

(x + y) + (x − y) = 38 + 4

which simplifies to 2x = 42, so x = 21. Now let's subtract the two equations instead:

x + y = 38
x − y = 4

Subtracting gives:

(x + y) − (x − y) = 38 − 4

which simplifies to 2y = 34, so y = 17. We've now found both original numbers.

Let's apply the same reasoning to Example 2. Call the two chosen numbers x and y. In step 2, we compute 2x + y and learn that 2x + y = 30. In step 3, we compute 2y + x and learn that it equals 27. So we know:

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

Let's add these two equations:

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

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

(2x + y) − (2y + x) = 30 − 27

which gives x − y = 3. Now that we know both the sum and the difference of x and y, we can use the same method as in Example 1 to find x = 11 and y = 8.

The key idea in these examples is learning how to transform one pair of simultaneous equations into another pair that has the same solution. When we want to disguise the original numbers, we keep applying operations until the information is sufficiently scrambled. To reverse the process, we choose operations that "unscramble" the information. The tools we use are:

a. Add equal quantities to equal quantities (x = 21, y = 17 became x + y = 38).
b. Subtract equal quantities from equal quantities (x = 21, y = 17 became x − y = 4).
c. Multiply both sides of an equation by the same nonzero number (for instance, multiplying both sides of 2y = 34 by 1/2).

Using these operations, we can transform one pair of simultaneous equations into another pair with the same solution. It's important to remember that since we have two unknowns, we must always work with two equations at each step. If I tell you my siblings' ages sum to 38, that alone isn't enough information to find both ages — I'd either need to tell you one age directly, or give you another clue, such as the difference between their ages.

The goal isn't to apply these operations randomly, but purposefully, working toward a solution. As we'll see in the next section, the specific numbers and the form of the equations tell us which operations to use. There's one more essential tool:

d. Substitution — replacing an expression in one equation with an equal expression taken from the other equation.

Example 3.

Lovasz has 5 marbles more than twice the number of marbles Tonio has. Together they have 107 marbles. How many marbles does Lovasz have?

Solution. First, we assign letters to the unknown quantities: let L represent Lovasz's marbles and T represent Tonio's marbles. The clue "Lovasz has 5 more than twice Tonio's" translates to L = 5 + 2T. The second clue — that together they have 107 marbles — gives us L + T = 107.

Since the first equation tells us L and 5 + 2T are equal, we can substitute 5 + 2T in place of L in the second equation:

5 + 2T + T = 107

From Chapter 1, we know how to solve a linear equation with one variable: combine like terms on the left, then subtract 5 from both sides to get 3T = 102, so T = 34. Tonio has 34 marbles.

To find L, we return to the first equation and substitute T = 34: L = 5 + 2(34), so L = 73. We can check this using the second equation: 73 + 34 = 107. ✓

Before using these operations to solve systems of equations, let's look at how this process appears graphically.

Example 4.

Consider the linear equations 3x + y = 7 and x + 3y = 5. Graph both equations on a coordinate plane and find the coordinates (x, y) of their point of intersection.

Solution. The slope of the first line is −3, and the slope of the second is −1/3. Since the slopes differ, the lines aren't parallel, so they must cross somewhere. Using a point from each line — for example, the y-intercepts (0, 7) for the first line and (0, 5/3) for the second — we can graph both lines, as shown in Figure 1.

The graph shows the lines intersecting at (2, 1). We can confirm this by checking that x = 2, y = 1 satisfies both equations:

3(2) + 1 = 7
2 + 3(1) = 5

In this example, we graphed the two linear relationships and read the coordinates of their intersection point, (2, 1), directly from the graph. Since this point lies on both lines, x = 2 and y = 1 satisfy both equations simultaneously. This is exactly what it means to "solve a pair of simultaneous equations."

Example 5.

Consider the equations x − 2y = 8 and 2x + 5y = 34. Graph both on the same coordinate grid, and read off the coordinates (x, y) of their point of intersection.

Solution. Again, the two lines have different slopes (1/2 and −2/5), so they're not parallel and must intersect somewhere. Graphing both equations (Figure 2), we find the intersection point at (12, 2). Since (12, 2) lies on both lines, x = 12 and y = 2 satisfy both equations. Because two nonparallel lines intersect at only one point, this is the only solution to the system.

What happens when the lines are parallel? Let's examine that case next.

Example 6.

Consider the equations 2x + 5y = 10 and 4x + 10y = 40. Graph each equation and look for a point of intersection.

These equations produce two different lines — they have different y-intercepts, (0, 2) and (0, 4) — but the same slope (−2/5), making them parallel. Since the lines never meet, there's no point of intersection, meaning there are no values of x and y that satisfy both equations at once. Figure 3 confirms this: the lines never cross.

Example 7.

Now consider the equations 2x + 5y = 20 and 4x + 10y = 40. These produce identical graphs, since both lines share the same slope and the same y-intercept.

If we rewrite both equations in slope-intercept form, they simplify to the same equation: y = −(2/5)x + 4. In this case, the system has infinitely many solutions, since every point on the line satisfies both equations. Notice that one original equation is simply a multiple of the other (divide both sides of the second equation by 2) — they're equivalent expressions. This pattern always occurs when a pair of simultaneous linear equations has more than one solution.

Summary: Given a pair of linear equations, there are three possible outcomes for simultaneous solutions:

1. Different slopes. The rate of change of y with respect to x differs between the two equations. Their graphs are lines with different slopes, so they aren't parallel and must intersect at exactly one point. The coordinates of this point form the unique solution.

1. Same slope, different intercepts. The rate of change of y with respect to x is identical for both equations, but they have different intercepts. This produces two parallel, distinct lines — meaning there is no solution to the system.

1. Same slope, same intercept. The rate of change of y with respect to x is identical, and the equations describe the same line. Every point on that line is a solution, so there are infinitely many solutions.

In short: if two lines have different slopes (are not parallel), they intersect at exactly one point. If two lines have the same slope (are parallel), then either there's no solution at all, or the lines are identical and there are infinitely many solutions.

Let's take a look at what happens graphically when we apply these operations to simultaneous equations. Start with the equations from Example 5: x − 2y = 8 and 2x + 5y = 34, graphed in Figure 2. If we add the two equations, we get 3x + 3y = 42. If we subtract the first equation from the second, we get x + 7y = 26. Adding the graphs of these two new equations to Figure 2 produces Figure 4, where the green and purple lines represent these additional equations. Notice that the