LE 10 Traditional Math: Elimination Method with Multiplication
Accelerated seventh grade and Math 8
This lesson continues with the elimination method in which one or both equations must be transformed to allow for cancellation of one of the variables. This causes some confusion at first, so I like to compare the process to finding a common denominator when adding fractions. Sometimes only one fraction needs to be changed, as is the case with 2/3 + 5/6. But other times both fractions need to be changed such as 7/9 + 2/5. Students generally find that connection useful.
Also students will sometimes plug the first answer into a transformed equation. While not wrong, it creates more work, so I emphasize to use the original equations to save time. Other times, students will multiply one side of the equation by a number, but not the other.
Warm-Ups.
1. Solve the system of equations. y + x = 15, x – y = 5. Answer: The first equation should be written as x + y = 15 to make it obvious how the elimination works. Students may have questions about this. x=10, y = 5.
2. Solve the system of equations. x + y = 7, x – y = 2; Answer: x = 9/2; y = 7 – 9/2; y = 14/2 – 9/2 = 5/2 Students may have questions on how to solve this. Prompts may include “If we substitute 9/2 for x then what must we do to isolate y? If we write 7 as 7/1 does what is the common denominator we need to do the subtraction?”
3. Express algebraically. A number is four more than another. (Let x represent one number, and y be four more than that number.) Answer: x = y +4;
4. Solve. 2/3 + 5/6 = ? Answer: 9/6
5. Solve 2/3 + 4/5 = ? Answer: 10/15 + 12/15 = 22/15
Problems 4 and 5 are included because of the similarity to transforming one or more equations to allow for elimination of a variable.
The Wind-Up and the Pitch: What if a Variable Can’t Be Eliminated? I write the following problem on the board and ask if anyone can solve it using the elimination method.
One of two things can happen. 1) Students will say it cannot be solved by elimination because none of the coefficients are the same. Or, 2) Someone will suggest that one of the equations be multiplied so that the coefficients are the same.
Of these two, option one is the more likely. If someone does know, I go with it, and have the student explain and follow-up. But let’s assume the more likely option occurs. I then refer to Problem 4 of the Warm-Ups. “How did we add the two fractions with different denominators?”
The common denominator is six for that problem so the first fraction was multiplied by 2/2. “Isn’t that similar to what’s going on here?”
Students will catch on, or I may supply more hints, but eventually they will see that the top equation or bottom equation can be multiplied by two.
“We want one pair of coefficients to cancel. If we multiply by two on the top, we get 6x, but I’d like that 6x to be negative, so I can add the two equations. So should it be two or something else?”
They see it is negative two and the resulting equation is -6x – 8y = -90. “Now we can add.” The result is y=6, x = 7. “When we plug in 6 for x, if you use the top equation use the original form—it’s easier. Or you can plug it in to the bottom equation.”
Examples:
2x + y = 5; 5x + 3y = 12; Answer; multiply top equation by -3. x = 3, y = -1
a-3b = 5; 3a + 2b = 4; Answer: multiply top equation by -3; a = 2, b = -1
3x + 2y = 12; 6x + 5y = 27; Answer, multiply top equation by -2, x=2, y =3
Transforming More Than One Equation. I now write the following equation on the board:
“This is similar to Problem 5 of the Warm-Ups; we had to change both denominators. So what do you think we need to do here?”
There will be some discussion. Some may want to multiply the top equation by -3 and the bottom by 2 (or vice versa). Others may want to multiply the top by -2 and the bottom by 5 (or vice versa.)
“Let’s go with top by -3 and bottom by 2 and see what gives.” They are to work along with me in their notebooks. We end up with:
Transformed equations are:
“We add the equations now and what do we get?” They should get -11y = -22. This yields y=2. “Where should we plug this in? Remember, plug it in to the original equation not the transformed one. Do this one by yourselves.” I walk around to check on their work. They should get x = 9, y = 2.
“Plug those values in to either equation to see if it checks.”\
Typical mistake: Students will multiply only the left hand side and forget to multiply the right hand side.
Examples: I work with them on the first two.
2x – 3y = -1; 3x – 4y = -3 Answer: Multiply top by -3, bottom by 2; x = -5, y = -3
2x + 4y =2; 3x + 5y = 2; Answer: Multiply top by -3, bottom by 2 or top by -5 and bottom by 4; x = -1, y = 1
3x – 2y = -5; -4x +3y= 8 Answer: Multiply top by 4, bottom by 3; x = 1, y = 4
Homework: Problems should be a mix of systems that don’t need transformations, and those that need one and two equations transformed. Problems with fractional answers should be included.