# Constants in Linear Equations

In this lesson we will become familiar and comfortable with the use of constants represented by letters. We will solve linear equations with these constants as if we knew the value of them. The unspecified constants are different to variables. Variables vary and constants remain constant! All these constants are doing is representing a singular value.

## Question 1a

Solve 4x + 7 = 27 for x.

Hint

Subtract both sides by 7

4x + 7 - 7 = 27 - 7

4x = 20

Divide both sides by 4,

4x / 4 = 20 / 4

x = 5

## Question 1b

Solve ax + b = c for x, where a, b and c are constants.

This may seem daunting! How can you solve something, with no numbers! We must remind ourselves what the symbols and letters in the equation mean, mathematics is a language. We are told a, b and c are constants, this means they could be something like 1 or 1028 or -4/3, but we just aren't told specifically what they are.

Begin to get comfortable with having these constants representing constant values.

It can help by comparing ax + b = c to the equation in 1a) 4x + 7 = 27,

4x + 7 = 27

ax + b = c

Now, if we imagine a, b and c representing the similar values in 4x + 7 = 27

As we move on to solve ax + b = c, think how we would operate if we knew what the constant values were, just as we did in a)

ax + b = c

First, we would subtract the b from both sides

ax + b - b = c - b

ax = c - b

[If it helps, think of c - b as a number as well, as we know, c and b are both numbers, so c - b would be a number as well]

Divide both sides a,

ax / a = (c - b) / a

x = {c - b} / a

There we have our solution, despite it maybe not looking like one. Remember, a, b and c are numbers, {c - b}/a would be a number as well.

Something extra: as we were “pretending” this was the same as question 1a), let’s try subbing in our a the same as 4, b the same as 7 and c the same as 27 into {c - b}/a;

{27 - 7}/4 =

20/4

5

Cool right!?!?

## Question 2

Solve {x - a}/ b + c = d for x with the constants a, b, c, and d.

Get into the mindset that as soon as you see “with the constants a....” immediately think of them as representing numbers. In the same way you might see x and start thinking about how you are going to solve it, see a constant represented by a letter and think that's a number I'll be working with, even though it's a letter!

If we are confused it might help to go over our inverse operations again,

So,

In order of what we would do to x,

1. Subtract a
2. Divide by b

So Inverse Operations, we do the opposite,

1. Subtract c
2. Multiply by b

So, for

{x - a}/ b + c = d

Subtract c,

{x - a}/ b = d - c

Multiply by b,

({x - a}/ b) b = (d - c)b

x - a = (d - c) b

x - a + a = (d - c) b + a

x = (d - c) b + a

## Question 3

Solve the equation 3(x + m) - n = 12 for x, with m and n being constants.

m and n are constants, so we imagine they are numbers.

How would we solve this if m and n were numbers? First thing would be to add the n to both sides,

3(x + m) - n = 12

3(x + m) - n + n = 12 + n

3(x + m) = 12 + n

Expand out the brackets,

3(x) + 3 (m) = 12 + n

Subtract the 3m from both sides (remember that 3m represents a constant number),

3x + 3m - 3m = 12 + n - 3m

3x = 12 + n - 3m

Divide by 3,

(3x) / 3 = (12 + n - 3m) / 3

x = {12 + n - 3m} / 3,

You can also correctly write the answer as,

x = 12/3 + n/3 - 3m/3

x = 4 + n/3 - m

Both are right.

For the next two questions, solve the equations for x. However, do not carry out any of the operations when solving. Leave all the numbers in the original equation as is. For example, if you had x + 1 = 3, leave your answer as x = 3 - 1 not x = 2. You can use the distributive property.

Doing this, we should still be able to see the numbers we have in the equation in our answer.

## Question 4a

Solve 5x + 2 = 2x + 30 for x

5x + 2 = 2x + 30,

First, we would subtract by 2x,

5x + 2 - 2x = 2x + 30 - 2x,

5x - 2x + 2 = 30,

We cannot carry out operations, but we can use the distributive property,

[ 5x - 2x = (5 - 2) x]

(5 - 2) x + 2 = 30,

Subtract 2 from both sides,

(5 - 2) x + 2 - 2 = 30 - 2

(5 - 2) x = 28

Divide both sides by (5 - 2),

((5 - 2) x) / (5 - 2) = 28 / (5 -2)

x = 28 / (5 - 2)

## Question 4b

Solve, for ax + b = cx + d with a, b, c, and d being constants

We will use the same idea as we had in a), with these constants. Think of the constants as numbers, just as we did in the previous question.

First, we would “get our x on one side”, so subtract by cx,

(ax + b) - cx = (cx + d) - cx

ax - cx + b = d

Think of the a and c multiplying the x as numbers, how would we operate around ax - cx? It would be the same as (a - c) times the x,

(just as it would be if it were 5x - 2x = (5 - 2) x = 3x),

So, write it as,

(a - c) x + b = d,

Now, subtract the b from both sides,

(a -c) x + b - b = d - b

(a -c) x = d - b

Reminding ourselves that a, b, c, and d are constants, divide both sides by (a - c),

((a - c)/x) / (a -c) = (d - b) / (a - c)

x = (d - b)/ (a - c),

There we have our solution to x, with the constants a, b, c, and d.

## Question 5

We have the expression 2a - b = c + 4d. Rearrange this equation four different ways to have an equation that shows the value of a, b, c, d each on their own.

2a - b = c + 4d

Subtract b from both sides,

2a - b - b = c + 4d - b

2a = c + 4d - b,

Divide both sides by 2,

a = {c + 4d - b} / 2

Now, for b=...

2a - b = c + 4d

Subtract 2a from both sides,

2a - b - 2a = c + 4d - 2a

-b = c + 4d - 2a

Divide both sides by -1,

-b / -1 = {c + 4d - 2a} / -1

b = {c + 4d -2a} / -1

For c =...

2a - b = c + 4d

Subtract 4d from both sides,

2a - b - 4d = c + 4d - 4d

2a - b - 4d = c

c = 2a - b - 4d

For d=...

2a - b = c + 4d

Subtract c from both sides,

2a - b -c = c + 4d - c

2a - b -c = 4d

Divide both sides by d,

{2a - b -c} / 4 = 4d / 4

{2a - b -c} / 4 = d

d = {2a - b -c} / 4