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Pearson Algebra 1 Common Core, 2011 View details

2. Solving Systems Using Substitution

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Exercise 55 Page 377

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

-3

Practice makes perfect

To solve the equation we have to isolate x 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.

5x+1=3x-5

▼

LHS-3x=RHS-3x

Evaluate

5x+1-3x=3x-5-3x

SubTerm

Subtract term

2x+1=-5

▼

LHS-1=RHS-1

Evaluate

2x+1-1=-5-1

SubTerm

Subtract term

2x=-6

Finally, we can fully isolate x by dividing both sides by 2.

2x=-6

▼

.LHS /2.=.RHS /2.

Evaluate

2x/2=-6/2

CalcQuot

Calculate quotient

x=-3

Therefore, x=-3 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.

5x+1=3x-5

Substitute

x= -3

5( -3)+1? =3( -3)-5

MultPosNeg

a(- b)=- a * b

-15+1? =-9-5

AddSubTerms

Add and subtract terms

-14=-14

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

Exercise 55 Page 377

Hints

Check the answer

Practice exercises

Asses your skill

Solving the Equation

Checking Our Answer