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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 57 Page 377

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

3/2

Practice makes perfect

To solve the equation we have to isolate k 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.

5k+7=3k+10

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LHS-3k=RHS-3k

LHS-3k=RHS-3k

5k+7-3k=3k+10-3k

Subtract term

2k+7=10

▼

LHS-7=RHS-7

LHS-7=RHS-7

2k+7-7=10-7

Subtract term

2k=3

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

2k=3

▼

.LHS /2.=.RHS /2.

.LHS /2.=.RHS /2.

2k/2=3/2

Calculate quotient

k=3/2

Therefore, k= 32 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.

5k+7=3k+10

k= 3/2

5( 3/2)+7? =3( 3/2)+10

Multiply

15/2+7? =9/2+10

a = 2* a/2

15/2+14/2? =9/2+20/2

Add fractions

29/2=29/2

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