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Core Connections: Course 3

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Core Connections: Course 3 View details

1. Section 6.5

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Exercise 38 Page 238

In a system of equations, if the left-hand sides of both equations are equal, the right-hand sides also are equal.

(a,b) = (-3, - 1/2)

Practice makes perfect

We want to solve the following system of equations using the Equal Values Method. a=12b+3 & (I) a = -2 b-4 & (II) In the given system, the left-hand sides of both equations are equal to each other — both are equal to a. That relationship means the right-hand sides of both equations are also equal to each other. The variable a can then be replaced by the right-hand sides, respectively. a= 12b+3 a = -2 b-4 ⇓ 12b+3 = -2 b-4 We know have an equation only in terms of b. Let's solve it!

12b+3 = -2 b-4

LHS+2b=RHS+2b

14b+3= -4

LHS-3=RHS-3

14b = -7

.LHS /14.=.RHS /14.

b = -7/14

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Simplify right-hand side

a/b=.a /7./.b /7.

b = -1/2

MoveNegNumToFrac

Put minus sign in front of fraction

b = -1/2

Now that we know the value of b, let's substitute it into Equation (II). We will get an equation only in terms of a. Then, let's solve our updated equation!

a = -2 b-4

b= -1/2

a = -2( -1/2 ) -4

MoveNegNumToDenom

Put minus sign in denominator

a = -2( 1/-2 ) -4

DenomMultFracToNumber

-2 * a/-2= a

a = 1 - 4

Subtract term

a = -3

Great work! We have one solution to our system of equations. (a, b) = ( -3, - 12)

Subchapter links

6.1.4 Using Rigid Transformations p.238