Core Connections Algebra 1, 2013
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Core Connections Algebra 1, 2013 View details
2. Section 1.2
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Exercise 40 Page 21

Practice makes perfect
a To solve the equation for x, we should isolate it on one side of the equation by performing inverse operations.
- 6 + 10x = x + 12
- 6 + 9x = 12
9x = 18
x=2
Let's now check this solution by substituting it into the equation. If it makes a true statement, then our solution is correct.
- 6 + 10x = x + 12
- 6 + 10( 2)? = 2+12
- 6 + 20? =2+12
14=14
The equation is true, so x=2 is a correct solution.
b Here, we can isolate x in a similar way by performing inverse operations.
10 + 5x = 2x - 11
10 + 3x = -11
3x=-21
x=-7
Now we check the solution by substituting it into the equation.
10 + 5x = 2x - 11
10 + 5( -7)? =2( -7) - 11
10-35? = -14-11
-25=-25
The equation is true, so x=-7 is the solution.
c Again, we will isolate x by performing inverse operations.
- 9x = - 3x + 18
- 6x=18
x=-3
Let's check the solution by substituting it into the equation.
- 9x = - 3x + 18
- 9( -3) ? = - 3( -3) + 18
27? = 9 + 18
27=27
The equation is true, so the solution is x=-3.
d To solve the equation, we will isolate x by performing operations.
2x - 9 = - 7x
-9 = -9x
1 =x
x=1
Let's now check this solution by substituting it into the equation. If it makes a true statement, then our solution is correct.
2x - 9 = - 7x
2( 1) - 9 ? = - 7( 1)
2-9? =-7
-7=-7
As we can see, x=1 is a correct solution.