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Big Ideas Math Integrated I, 2016

BI

Big Ideas Math Integrated I, 2016 View details

2. Solving Systems of Linear Equations by Substitution

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Exercise 10 Page 227

Does either of the equations have an isolated variable in it?

(1, -3)

Practice makes perfect

When solving a system of equations using the Substitution Method, there are three steps.

Isolate a variable in one of the equations.
Substitute the expression for that variable into the other equation and solve.
Substitute this solution into one of the equations and solve for the value of the other variable.

For this exercise, y is already isolated in one equation, so we can skip straight to solving!

6x-9=y & (I) y=-3x & (II)

(I):y= -3x

6x-9= -3x y=-3x

(I):LHS-6x=RHS-6x

-9=-9x y=-3x

(I): Rearrange equation

-9x=-9 y=-3x

(I): .LHS /(-9).=.RHS /(-9).

x=1 y=-3x

Great! Now, to find the value of y, we need to substitute x=1 into either one of the equations in the given system. Let's use the second equation.

x=1 y=-3x

(II):x= 1

x=1 y=-3 (1)

(II): Identity Property of Multiplication

x=1 y=-3

The solution, or point of intersection, to this system of equations is the point (1, -3).

Checking Our Answer

To check our answer, we will substitute our solution into both equations. If doing so results in true statements, then our solution is correct.

6x-9=y & (I) y=-3x & (II)

(I), (II): x= 1, y= -3

6 (1) -9? = -3 -3? =-3 (1)

(I), (II): Identity Property of Multiplication

6-9? =-3 -3=-3 ✓

(I): Subtract term

-3=-3 ✓ -3=-3 ✓

Because both equations are true statements, we know that our solution is correct.

Subchapter links

Communicate Your Answer p.223

Monitoring Progress p.224-226

Exercises p.227-228