| Solving Linear Equations |
LINEAR EQUATIONS
This is an Equation
Now that we've learned the syntax, order and grammar rules of the language known as algebra, we can use it to write sentences or equations to describe a myriad of situations. Equations are used in Physics, Chemistry, Biology, Computer Programing, Banking, --- you name it. Why, I used a simple algebraic equation one morning to calculate the mix of water and 10% cream I needed to supply enough whole milk (3.5%) for my bowl of breakfast cereal. The timing of traffic light changes is determined by equations based on statistical analysis of traffic flow.
An algebraic equation works exactly the same way balanced scales do, with the equal sign replacing the pivot point of the scales. Whatever you do to one side of the equation must be done to the other side in order to maintain equality or balance. Those are the operational words. We must maintain equality at all times -- why do you think we call it an EQUATION?? We would call it something else if it was something else -- but it's not. There are other types of statements in algebra such as inequalities -- (coming up in lesson 7) -- and functions -- which are equations but different and the same. In this lesson, we concentrate on the equality relation.
A binary relation is a relation between two terms or expressions. For example, a = 3 is a binary relation statement. The relation symbol is = (equality) and the "binary" part -- binary meaning 2 things are involved -- are the a and the 3. Here's another binary statement: 4(x + 1) > 2. This relation symbol is greater than and its counterpart less than looks like this < . With just 2 elements in the relation, the only other possibilities are greater than or equal to (
) or less than or equal to (
).
There are three binary relations in mathematics.
They are: ..... equal to ( = ), ..... greater than ( > )..... and ..... less than ( < ).
We begin with equations: statements of equality. Then we'll investigate the less than and greater than relationships -- the inequalities in subsequent algebra lessons.
A linear equation is one in which all the variables are to the first power. Should we decide to graph a linear equation, we would get a line. To solve a linear equation, we simply isolate the variable for which we wish to solve, and perform the correct algebraic operations to get a statement about its value.
Whatever we do to one side of the equation, we must do to the other.
Example: Solve x + 7 = 10 . To solve, we must find the value of x that makes this true.
Our last statement in the solution will be x = some number.
If we could take away the 7 from the left side, we'd be left with what we want, so that's what we do. Since it's an equation and we must always maintain equality, we have to
subtract (take away) 7 from both sides to get: x + 7 – 7 = 10 – 7 which becomes x = 3.
To check our solution, we replace x with 3 in the original equation.
Since 3 + 7 = 10, we have solved the question correctly.
Notice how the "7" seemed to jump over the equal sign and in doing so, changed from positive to negative.
Had our equation been x – 7 = 10, we would add 7 to both sides. We would get:
x – 7 + 7 = 10 + 7 which becomes x = 17
and when we check, we get 17 – 7 = 10.
Again, the "– 7" seemed to jump over the equal sign, but this time, it changed from negative to positive. To shorten the solving process, we use a technique called transposing.
.
Transposing: shortcut to solving equations
We often refer to adding and subtracting terms in an equation as transposing.
We use it to move algebraic terms from one side of the equal sign to the other -- by changing their sign.
Here's an example.
Solve: 5x – 1 = 3x + 7
We'll move the 3x over to the left side, and we'll move the – 1 over to the right side of the =.
5x – 1 = 3x + 7 ..... becomes ..... 5x – 3x = 7 + 1
we collect like terms to get:..... 2x = 8 ..... so we know x = 4.
Notice how 3x became – 3x, and how – 1 became + 1 when they showed up on the opposite side of the equation. Transposing only works for addition and subtraction. In the last step, we divided both sides by 2 so that we could state the value of "one" x -- as the solution demands.
Here's a more complicated example. First we'll simplify the expressions in the equation, then we'll collect like terms and only then will we transpose terms to organize the setup for the solution.
Solve for x if 3(3x – 1) = 4(x – 3) – 1
Step 1: remove brackets ..... ..... ..... ..... 9x – 3 = 4x – 12 – 1
Step 2: collect terms (– 12 and – 1) ..... .. 9x – 3 = 4x – 13
Step 3: transpose – 3 and 4x. ... ..... ..... 9x – 4x = – 13 + 3
Step 4: collect terms ..... ..... ..... ..... ... 5x = – 10 ... ..so ... ..x = – 2 by division.
When transposing, we must change the sign of any term that crosses the equal sign.
Notice that the – 3 became + 3 and the 4x became – 4x when they were transposed.
Both methods give the same results since they are identical, but transposing is more efficient.
We use linear equations to solve word problems that involve variables to the first power. Some of these problems are best solved using a table or diagram in conjunction with equations. Others can be handled with just an equation preceded by a " let " statement, like let x = the amount of money Joe has. This is known as defining the variable. More on solving word problems can be found in other Algebra MathRoom lessons.
Practice
Solve these equations for all possible values of the variable:
1) 3( y – 1) – 1 = 2 – 5( y + 5)
2) 3x – 4(x + 6) = 2[x – 3(5 – x)]
3) 3(4x – 1) = 9 – 4x
4) 5x – 3(x + 1) = 2(3x + 5)
5) 15w – 4 = 8w + 31
.
Solutions
| 1) 3(y – 1) – 1 = 2 – 5(y + 5) 3y – 3 – 1 = 2 – 5y – 25 8y = – 19 . y = – 19 / 8 |
2) 3x – 4(x + 6) = 2[x – 3(5 – x)] 3x – 4x – 24 = 2[ x – 15 + 3x ] – x – 24 = 2( 4x – 15 ) = 8x – 30 – 9x = – 6 . x = 2 / 3 |
3) 3(4x – 1) = 9 – 4x 12x – 3 = 9 – 4x 16w = 12 . w = ¾ |
| 4) 5x – 3(x + 1) = 2(3x + 5) 5x – 3x – 3 = 6x + 10 – 4x = 13 . x = – 13/4 |
5) 15w – 4 = 8w + 31 7w = 35 . w = 5 |
(all content © MathRoom Learning Service; 2004 - ).