Big Ideas Math: Modeling Real Life, Grade 8
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3. Solving Systems of Linear Equations by Elimination
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Exercise 14 Page 214

Which method can we use to solve each system of equations?

See solution.

Practice makes perfect

We have four system of linear equations and we want to decide which of them does not belong with the other three.

We can see that in three cases we have a pair of variables with opposite coefficients.

Therefore, we can use elimination to solve these three systems. This means that the third system does not belong with the other three. Let's try to solve the first system. Since the y-terms have opposite coefficients we will solve this system with elimination.
3x+3y=3 & (I) 2x-3y=7 & (II)
3x+3y=3 2x-3y+ 3x+3y=7+ 3
3x+3y=3 5x=10
3x+3y=3 x=2
Now we will substitute 2 for x into Equation 1.
3x+3y=3 x=2
3( 2)+3y=3 x=2
6+3y=3 x=2
3y=-3 x=2
y=-1 x=2
The solution to the system is (2,-1). We can solve the second and the last systems in a similar way. Now let's try to solve the third system. Since neither pair of variables has opposite coefficients, we cannot use elimination straight away.
We can see that the y-terms have opposite signs. However, before we can add the equations we need to multiply Equation 1 by 2 and Equation 2 by 3. By doing this, we will end with opposite coefficients.
2x+3y=11 & (I) 3x-2y=10 & (II)
2(2x+3y)= 2(11) 3x-2y=10
2(2x)+2(3y)=2(11) 3x-2y=10
4x+6y=22 3x-2y=10
4x+6y=22 3(3x)- 3(2y)= 3(10)
4x+6y=22 9x-6y=30
Now the y-terms have opposite coefficients. This means that we can add Equation 2 to Equation 1 to eliminate the y-terms.
4x+6y=22 9x-6y=30
4x+6y+ 9x-6y=22+ 30 9x-6y=30
13x=52 9x-6y=30
x=4 9x-6y=30
Finally we will substitute 4 for x into Equation 2.
x=4 9x-6y=30
x=4 9( 4)-6y=30
x=4 36-6y=30
x=4 -6y=-6
x=4 y=1
The solution to this system is (4,1).