Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
4. Solving Equations With Variables on Both Sides
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Exercise 55 Page 108

Start with an expression for the left-hand side and calculate its value when x equals any fraction.

Example Solution: 2x+9=x+9 12

Practice makes perfect
We are going to write an equation which has a fraction as a solution and contains the variable x on both sides. We can start off with any linear expression for the left-hand side. 2x+9 Now let's choose that our solution will be x= 12. By substituting 12 for x into the equation, we can determine what the expression equals.
2x+9
2( 1/2)+9
1+9
10
This means that the expression 2x+9 needs to be equal to 10. 2x+9=10 Let's find another expression that equals 10 when x= 12. The simplest way to do this is by writing an expression where a number n added to x equals 10. x+n=10 To find n, we should substitute x= 12 for x.
x+n=10
1/2+n=10
â–Ľ
LHS-1/2=RHS-1/2
n=10-1/2
n=20/2-1/2
n=19/2
n=18+1/2
n=18/2+1/2
n=9+1/2
n=9 12
We found another expression that equals 10 when x= 12. x+9 12 Since both expressions equal 10 when x= 12, we can set them equal to one another. This will be our equation. 2x+9= x+9 12 Let's check our solution just to make sure. To do this we will solve our equation using the Subtraction Property of Equality.
2x+9=x+9 12
â–Ľ
LHS-9=RHS-9
2x=x+9 12-9
2x=x+19/2-9
2x=x+19/2-18/2
2x=x+1/2
x=1/2
As we can see, x is a fraction and our solution is correct. Remember that this is just one example of a possible solution for this problem.