Exercise 2 Page 48

We are asked to consider the following equation. cx-a=b First, recall that two equations are equivalent if we can get one from the other using different Properties of Equality. Keeping in mind this information, we will consider each of the listed equations separately.

cx-a+b=2b

In this equation, we can see that on the left-hand side we have an extra b. Let's add b to both sides of the original equation to see if we will get the considered equation. Since it resulted in the same equation, we know that these two equations are equivalent.

cx-a+b=2b ⇔ cx-a=b

0=cx-a+b

This equation stands out with 0 on the left-hand side, so let's rearrange the original equation so that it has 0 on one of the sides. We can do it by subtracting b from both sides.

cx-a=b

SubEqn

LHS-b=RHS-b

cx-a-b=0

RearrangeEqn

Rearrange equation

0=cx-a-b

We did not get the same equation, as b is subtracted form the right-hand side, not added. Therefore, the equation is not equivalent to the original.

2cx-2a=b/2

Here, we can see that the left-hand side has been multiplied by 2. Therefore, let's multiply the original equation by 2 to see if both sides match.

cx-a=b

MultEqn

LHS * 2=RHS* 2

2cx-2a=2b

These two equations are not equivalent, as the right-hand sides are different.

x-a=b/c

In this equation, we see that instead of cx we have x, so let's divide the original equation by c.

cx-a=b

DivEqn

.LHS /c.=.RHS /c.

cx-a/c=b/c

WriteDiffFrac

Write as a difference of fractions

cx/c-a/c=b/c

SimpQuot

Simplify quotient

x-a/c=b/c

Comparing these two equations, we see that they are not equivalent.

x=a+b/c

In this equation x is isolated. Therefore, we are going isolate x in the original one and see if the right-hand sides are the same.

cx-a=b

AddEqn

LHS+a=RHS+a

cx=b+a

CommutativePropAdd

Commutative Property of Addition

cx=a+b

DivEqn

.LHS /c.=.RHS /c.

x=a+b/c

We got the same equation, so they are equivalent. x=a+b/c ⇔ cx-a=b

b+a=cx

Here we have b+a on the left-hand side, so let's add a to the original equation.

cx-a=b

AddEqn

LHS+a=RHS+a

cx-a+a=b+a

AddTerms

Add terms

cx=b+a

RearrangeEqn

Rearrange equation

b+a=cx

We got the same equation, so they are equivalent. b+a=cx ⇔ cx-a=b