Algebra Know-It-ALL

Chapter 12

Question 12-1

Suppose a is a constant and x is a variable for which we want to solve. We see the following
equation:

x− 2 a= 5 a

How can we get an equation with x alone on the left side of the equals sign and a multiple of
a alone on the right?

Answer 12-1

We can morph the equation as follows:

x− 2 a= 5 a

(x− 2 a)+ 2 a= 5a + 2a

x= 7 a

Question 12-2

Suppose b is a constant and z is a variable for which we want to solve. We see:

z+ 3 b=b/2

How can we get an equation with z alone on the left side of the equals sign and a multiple of
b alone on the right?

Answer 12-2

We can morph the equation as follows:

z+ 3 b=b/2

2(z+ 3 b)= 2(b/2)

2 z+ 6 b=b

(2z+ 6 b)− 6 b=b− 6 b

2 z=− 5 b

z= (−5/2)b

Question 12-3

When we divide both sides of a first-degree equation by any expression containing the vari-
able, there’s a hidden risk. What is that risk?

Answer 12-3

The expression containing the variable must not be equal to 0, or we’ll run into trouble of some
sort. The fact that the expression contains an unknown means that we can’t be sure it’s nonzero
until we know the solution to the equation! When trying to solve a first-degree equation, there-
fore, it’s best to avoid dividing through by any expression that contains the variable.

Part Two 311