Linear Equations

An Equation is a statement that says two algebraic expressions are equal. Equations may be true or false. We identify three types of equations based on whether or not replacements exist for the variable that will make the quation true.

1. Identities: ALL replacements for the variable make the equation true.

Example: x + 1 = 1 + x (True for all real numbers) Solution Set: {x | x is any real number}

Try this example: 2(1 – x) = –2x + 2

2. Contradictions: NO replacements for the variable make the equation true.

Example: x + 1 = 2 + x (True for NO real numbers) Solution Set: Empty Set { }=Æ .

Try this example: 2x + 3 = 2(x + 1) + 3

3. Conditional Equations: SOME replacements for the variable make the equation true.

Example: x + 1 = 3 (True for one real number; 2 only) Solution Set: {x | x = 2}

 Try this example: 2(x – 1) = x + 1

 

An equation is referred to as a linear equation, if the variable we have to solve for has no powers and does not appear in the denominator of a fraction.

For example 2(x – 5) – 15 = 7x – 20 is a linear equation.

In solving any equation our goal is to find the value that could replace the variable in such a way that the equation remains true. If we replace x with 7 in the linear equation above the two sides of the equation won’t be equal to each other: 2(7 – 5) – 15 is –11, while 7(7) – 20 is 29. But if we replace x with –1 the two sides of the equation will give us the same answer: 2(–1 – 5) – 15 is –27 and 7(-1) – 20 is also –27. So we say that –1 is a solution to our equation. But how can we find the solution the a linear equation?

Think of an equation as a scale or balance, the expressions on each side are equal so they balance each other. If we want to we can double both sides and they will still balance, because we have increased them equally. By the same token we can add or subtract the same amount on both sides and they will still be equal. In fact, the two sides of an equation are going to be equal no matter what operation we perform as long as it is done to both sides of the equation. This is the fact we are going to use to solve a linear equation.

 

Remember that our goal in solving a linear equation is to find x (or whatever variable is used). That means we want to manipulate the equation until we get to an expression that says: x = C (where C is a number or an expression that does not have an x in it). In the above example that expression is going to be x = –1.

Lets examine our equation:

Example 1) Solve 2(x – 5) – 15 = 7x – 20

First thing we’ll do is try to simplify each side of the equation as much as possible. Remember that we may combine terms that are on the same side of the equation, but not those that are on the opposite sides.

2x – 10 –15 = 7x – 20

2x – 25 = 7x – 20

Notice that the above expression has variables on both sides of the equation, and we want to end up with x = C and C can’t have x’s in it. So we need to eliminate x’s from one side of this equation. It doesn’t really matter which side, but for the sake of consistency I will always try to keep the x’s on the left side. This means that I want to eliminate the 7x from the right hand side. We can do that by subtracting a 7x from both sides of the equation. Subtracting 7x cancels the 7x on the right hand side and since I am going to also subtract it from the left hand side the equality is maintained.

2x – 25 – 7x = 7x – 20 – 7x

–5x – 25 = –20

Now we have to eliminate the –25 from the left hand side. And we can do that by adding 25 to both sides of this equation.

–5x – 25 + 25 = –20 + 25

–5x = 5

And finally to undo the –5 that is multiplied by x we can divide both sides of this equation by a –5.

  So we get x = –1.

NOW WHAT IF OUR EQUATION INVOLVES FRACTIONS?

Example 2)

We’ll start the problem the same way as the previous one, by distributing and simplifying as much as possible.

(LCD of 3, 5 and 20 is 60)

Now since this is an equation and we are allowed to perform any arithmetic operation as long as we do it on both sides, we can attempt to eliminate the denominators. We achieve that by multiplying both sides of the equation by the common denominator of the fractions on both sides of the equation. in this case 60.

240x – 200 = 120 + 24x – 3

240x – 200 = 117 + 24x

We start by eliminating the x’s on the right hand side by subtracting 24x.

240x – 200 – 24x = 117 + 24x – 24x

216x – 200 = 117

Then eliminate the –200 by adding 200 on both sides,

216x – 200 + 200 = 117 + 200

216x = 317

And then divide by 216 on both sides to undo the 216 multiplied by x.

So x = 317 / 216.

SOMETIMES WE MAY HAVE LINEAR EQUATIONS THAT INVOLVE MORE THAN ONE VARIABLE:

 

Example 3) Solve 4at + 5y = 2t + 9y for t.

Since we are solving this equation for t, we need to make sure that we end up with t = C where C is either a number or in this case an expression that has no t’s in it. So we start by eliminating the t’s on the right hand side of this equation by subtracting 2t from both sides.

4at + 5y – 2t = 2t + 9y – 2t

4at + 5y – 2t = 9y

Next we need to eliminate the 5y from the left hand side (because it has no t’s). We can do this by subtracting 5y from both sides.

4at + 5y – 2t – 5y = 9y – 5y

4at – 2t = 4y

Up to this point we have proceeded exactly the same as the previous problems, but now we have a new situation on the left hand side. We have more than one t and they can’t be combined. Remember that our goal is to be left with a single t on the left side. In situations like this, we can factor the expression and that will allow us to write our multiple terms as one term.

2t(2a – 1) = 4y

Now we can isolate the variable t by dividing both sides by 2(2a – 1):

Which after reducing gives us: With restriction:

Applications and Word Problems