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Pearson Algebra 1 Common Core, 2011

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Pearson Algebra 1 Common Core, 2011 View details

2. Solving Systems Using Substitution

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Exercise 56 Page 377

Isolate c on one side of the equation using the Properties of Equality.

2

Practice makes perfect

To solve the equation we have to isolate c on one side.

Solving the Equation

We will begin by moving all variable terms to the left-hand side and all constants to the right-hand side.

4c-7=- c+3

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LHS+c=RHS+c

LHS+c=RHS+c

4c-7+c=- c+3+c

Add terms

5c-7=3

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LHS+7=RHS+7

LHS+7=RHS+7

5c-7+7=3+7

Add terms

5c=10

Finally, we can fully isolate c by dividing both sides by 5.

5c=10

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.LHS /5.=.RHS /5.

.LHS /5.=.RHS /5.

5c/5=10/5

Calculate quotient

c=2

Therefore, c=2 is the solution to the equation.

Checking Our Answer

To check that our answer is correct, we can substitute our solution back into the given equation.

4c-7=- c+3

c= 2

4( 2)-7? =-( 2)+3

Multiply

8-7? =-2+3

Add and subtract terms

1=1

Because both sides are equal, we know that our solution is correct.

Subchapter links

Lesson Check p.375

Practice and Problem-Solving Exercises p.375-377

Standardized Test Prep p.377

Mixed Review p.377