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Big Ideas Math Integrated I, 2016

BI

Big Ideas Math Integrated I, 2016 View details

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Exercise 3 Page 47

Isolate x on one side of the equation.

x=-1/5

Practice makes perfect

We need to isolate x on one side to solve the equation. By using the Subtraction Property of Equality twice, we can collect all the x-values on the left-hand side and all of the constant terms on the other side of the equation.

11x+1=-1+x

LHS-x=RHS-x

11x+1-x=-1+x-x

Subtract terms

10x+1=-1

LHS-1=RHS-1

10x+1-1=-1-1

Subtract terms

10x=- 2

Now, we need to isolate x by dividing both sides by 10 by using the Division Property of Equality.

10x=- 2

.LHS /10.=.RHS /10.

10x/10=- 2/10

Calculate quotient

x=-1/5

The solution is x=-1/5. We can check our solution by substituting it into the original equation.

11x+1=-1+x

x= -1/5

11*( -1/5) +1 ? = -1 + ( -1/5)

▼

Simplify

MoveNegFracToNum

Put minus sign in numerator

11*(-1/5) +1 ? = -1 + (-1/5)

MoveLeftFacToNum

a*b/c= a* b/c

(-11/5) +1 ? = -1 + (-1/5)

Rewrite 1 as 5/5

(-11/5) +5/5 ? = -5/5 + (-1/5)

Remove parentheses

-11/5 +5/5 ? = -5/5 + -1/5

Add fractions

-6/5=-6/5 ✓

Since the left-hand side is equal to the right-hand side, our solution is correct.

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Exercises p.47