In Activity 2, you finished by noticing that:

• A linear system with 1 point of intersection has linear relations with different slope values.
• A linear system with 0 points of intersection has linear relations with the same slope value but different y-intercepts.
• A linear system with every point intersecting has two linear relations with the same slope value and the same y-intercept. They are the same line.

The information about slope and y-intercept is in every equation, but, as you have already seen, not all equations display the information in the same way. Using the slope y-intercept form, you can easily read slope and y-intercept values, while in the standard and alternate forms, some additional calculation is usually required.

In this activity you will be using the alternate form of the linear relation, Ax + By = C. Recall that in this form, A, B and C are integers(definition:any one of the numbers …,-3, -2, -1, 0, 1, 2, 3, …).

To summarize what you were explored...

Equivalent equations:

• to create an equivalent equation, you can multiply or divide each term in the linear relation by the same value;
• equivalent equations have the same key features, x-intercept, y-intercept and slope.

Combining equations:

• when two linear relations are combined (added or subtracted) to produce a third linear relation and it is graphed with the two original equations, all lines will pass through the same point on the graph. The point of intersection is the same.

As you move forward in this activity, you will use the concepts of equivalent equations and combining equations to determine the point of intersection for a linear system.

In the previous activity, Activity 2, you learned to solve a linear system using the algebraic method of substitution. You rewrote one linear relation to isolate a variable and create an expression for the variable. You substituted that expression into the other linear relation creating an equation with only one variable. By eliminating a variable, you made it possible to solve the equation for the remaining variable.

In this activity, you will be solving a system of linear equations using an algebraic method called elimination. You will strategically select equivalent equations then add or subtract to eliminate one of the variables.

Solving a System by Elimination

During the Minds On task, you looked at equivalent equations for a linear relation and noticed that you can create an equivalent equation by multiplying (or dividing) each term by the same value. With equivalent equations, every point on the line is a point of intersection with the other line.

You also looked at equations that were not equivalent; their slopes were different and they had one point of intersection. Additional equations were added to the same graph and they also passed through the same point of intersection. You noticed that the coefficients and constant value of each new equation were a result of adding or subtracting the like terms of the two initial equations.

You have seen that when the solution to a linear system is a point, it can be written as an ordered pair (e.g. (3, 1) ). The solution can also be seen on a graph as the intersection of the vertical line for the x value (e.g. x = 3) and a horizontal line for the y value (e.g. y = 1). Think about how you might use equivalent equations along with adding or subtracting terms to yield a horizontal or vertical line.

Example 1

How do I know that a point of intersection exists? What information is available from each equation?

Find the point of intersection of the two linear relations:

1. x - y = 1
2. x + y = 7

How can you check that there will be one point of intersection?

Answer

What is the value of the slope for equation 1?

Answer

What is the value of the slope for equation 1?

Answer

If I recall that the value for the slope is:

Otherwise, I can use the points (7, 0) and (0, 7) and calculate:

If I rearrange equation 2 as I did for equation 1, I get y = -x + 7, and I can confirm that the slope is -1.

Since the slopes of the two lines are different, a point of intersection must exist.

To solve an equation through the algebraic process of elimination, you will learn to add or subtract the coefficients in the equations in a way that the result will be the equation of a vertical line or a horizontal line. This means that through addition or subtraction, the coefficient of either the x term or the y term will be 0 and that term is eliminated.

The x term would be eliminated if the coefficients of x in the two equations were opposite integers and you added the like terms to yield 0x.  Opposite integers are two integers that are the same distance away from 0 and on opposite sides of the number line (e.g.: 2 and -2, 6 and -6).

The x term would also be eliminated if the coefficients in the two equations were the same integers and you subtracted the like terms to yield 0x.

What are the coefficients of the x and y terms in each equation? Are they the same integer? Are they opposite integers?

Find the point of intersection of the two linear relations:

x - y = 1
x + y = 7

What are the coefficients of x and y in each equation?

Answer

Should I add or subtract the coefficients and the constant term?

Answer

If you use the two equations, x – y = 1 and x + y = 7, you could rearrange both equations to equal 0.

x – y = 1 becomes x – y – 1 = 0 and

x + y = 7 becomes x + y – 7 = 0.

Since both equations equal the same value, 0, they must be equal so you could write

x – y – 1 = x + y – 7

And now, using your equation-solving skills, you can rearrange the equation to combine like terms.

x – x – y – y -1 + 1 = x – x + y – y – 7 + 1

Notice that the bolded portions are the net result of adding/subtracting the same term on both sides of the equation. We are trying to achieve having variables on the left side of the equals sign and constants on the right side.

Notice that like terms are being combined. Here you can see that the resulting equation would be -2y =  – 6, which can be simplified to y = 3.

Mathematicians use the idea of equations being equal to the same value (e.g. 0) to combine like terms, but they work with the equations in a simpler way. 

In the example above, the coefficients were either the same or opposite values. You can see that if x was eliminated by subtraction first, the value for y, y = 3, would be used to determine the value for x.

In the next example, neither adding nor subtracting will eliminate a term because the coefficients of x and y are neither the same integer nor opposite integers. In this situation, we will introduce an equivalent equation to create the situation that will allow for one term to be eliminated.

Example 2

Solve the system:

4x + 3y = 4
8x - y = 1

Neither x nor y have the same or opposite coefficients so I need to use my skills and create an equivalent equation that will allow me to add or subtract to eliminate a variable.

Read each solution, eliminate x and eliminate the y variable first, to see the difference in the process but the identical answer at the end.

Possible Solution 1 - Eliminate x

Possible Solution 2 - Eliminate the y variable first

 

Both methods, eliminate x first or eliminate y first, produced the same result through a different but equivalent pathway.  Regardless of the choice of pathway, you probably noticed that the same steps were followed using the different values.

The previous example would have been difficult to solve graphically by hand because one of the coordinates of the solution, graphically the point of intersection, is not an integer value. It may take a number of tries to approximate the correct value for x from the graph. An algebraic solution is more timely and efficient.

Different Solutions in Linear Systems

You have been working with systems of linear equations that have solutions or a point of intersection.  You recognized earlier that any two lines with different slopes will intersect and it is possible to solve the system using one of the three methods that you have learned. You could solve by graphing, solve algebraically using substitution, or solve algebraically using elimination.

Some linear systems will not yield one point of intersection. A linear system with two parallel lines does not have a point of intersection while a linear system with two equivalent equations or the same line will have an infinite number of points in common.

You already know what they look like graphically but it is also useful to know what these look like algebraically. The three situations below show the three different outcomes.

 

 

 

You have explored the process of solving a linear system by using equivalent equations to create another equation that is either the horizontal line or the vertical line that passes through the point of intersection. You eliminated either the x term or the y term to produce the new equation. From there, you used the information to determine the point on one line and then verify that the same point was also on the second line.

Congratulations, you have completed Unit 3, Activity 3. You may move on to Unit 3, Activity 4.