Solving Equations - 1

Algebraic Equations: Solving it with Algebra

this is an equation An Equation is an algebraic statement of Equality
between knowns (constants) and unknowns (variables).

An algebraic equation behaves exactly the same way a seesaw does or balanced scales do, with the equal sign replacing the pivot point of the seesaw or scales. Whatever we do to one side of the equation, we do to the other side so that we 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.

When we solve an equation, our job is to find the number value(s) of the unknown or variable that makes the equation a true statement.

 The solution of an equation is a number value for the unknown (variable) that satisfies the equation. The solution is also called the "root" of the equation. It makes the equation a true statement.

The only unbreakable rule of equations is:

Whatever we do to one side of the equation, we must do to the other.

Let's solve the equation in the graphic above.

We have 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 discussed later in this lesson.

In the examples above, there is exactly one solution for each equation. x = 3 is the only number value in the entire universe that makes x + 7 = 10 a true statement. Similarly, only 17 - 7 = 10 so the only solution for the 2nd equation is x = 17.

Some equations have more than one solution.
For instance, x ² = 36 has 2 roots, because 6 ² = 36 and ( – 6)² = 36. Both 6 and – 6 make this statement true, so they are both solutions. However, had this equation represented the area of a square that is x units per side, we could not accept – 6 as a solution since we can't draw a square with sides that measure – 6 units. If the equation we're solving comes from a word problem, we must make sure our solutions fit the reality of the problem. We cannot represent dimensions with negative numbers.

In the first example, we had x + 7 = 10 so we subtracted 7 from both sides.
In the 2nd example, we had x - 7 = 10 so we added 7 to both sides.
If we had 8x = 24, we would divide both sides by 8 to get x = 3.
If we have to solve , we would multiply both sides by 3 to get x = 3 × 9 = 27.

 To solve an equation, we perform the INVERSE of the operations in it.

When we see subtraction in the equation, we add. When we see division, we multiply. The only rule is no prejudice!! What we do to one side, we do to the other -- ALWAYS!!

All our equations so far have required only 1 step or operation to solve. Now we look at transposing -- a faster way to perform those operations and we look at solving equations that require more than one step because more than one operation was performed on the variable or unknown.

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 divide by 2 to get 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.

Equations Solved with More Than One Step

Our goal when solving equations is to isolate the variable. We want the unknown term to be all by itself on one side of the equation and we want a number on the other side. To accomplish this, we undo or invert any operations indicated in the equation. Once we have the unknown term equals a precise number value, we have our solution.

Examples

1) Solve this equation: 5x – 7 = 33

when we transpose the7 to the right side, we get:
5x = 33 + 7
Now we have 5x = 40
so we divide both sides by 5 to get x = 8.

To check, we verify if the original statement is true. 5(8) – 7 does = 33 so we're done.

2) Solve this equation: 3x + 4 = 37

when we transpose the 4 from left to right, we get: 3x = 37 – 4 or 3x = 33
Now we see multiplication by 3, so we divide both sides by 3 to get x = 11.

When we substitute 11 for x in the original statement we see it is true. 3(11) + 4 does = 37.

3) Solve this equation: Here we see addition and division (remember, fractions mean division), so first,
we transpose the 9 and then we multiply by 5 to find x. We get: This says that x divided by 5 equals 6, so x must be 5 × 6 or 30.

When we check, we get (30 ÷ 5) + 9 = 6 + 9 = 15, so our solution is correct.

We mustn't be fooled into believing that all equations begin with all the variables on one side or the other. There are times we have to do some organizing before we have all the variables together like this:

Example: Solve 5m – 3 = 2m + 6

Here, we have variables and constants on both sides of the equal sign.
First, we transpose the – 3 and the 2m then we collect terms:

5m – 2m = 6 + 3 ..... which is ..... 3m = 9

Now, we divide both sides by 3 to get:

m = 3

When we check our solution in the original equation, we get:

5(3) – 3 = 2(3) + 6 which says 15 – 3 = 6 + 6 or 12 = 12.

Our solution is correct.

 To solve an equation with more than 1 variable term, collect them all on one side, then solve.

.

Now get a pencil, an eraser and a note book, copy the questions,
do the practice exercise(s), then check your work with the solutions.
If you get stuck, review the examples in the lesson, then try again. .

Practice Exercises

1) Describe the single operation we do to both sides to solve, then solve mentally.

 a) x – 12 = 22 b) x + 19 = 229 c) 3x = 45 d) 2) State the two operations we use to solve these,then find the solution.

 a) 2x – 12 = 22 b) 7x + 19 = 229 c) 3x – 3 = 48 d) ¼ x – 3 = 1

3) Solve these equations using transposing. Show all your work.

 a) 15x – 2 = 19 – 6x b) 4t + 5 = 35 – t c) 13x – 20 = 40 + x d) 9m – 2 = 4m + 23 Solutions

1) Describe the single operation we do to both sides to solve, then solve mentally.

 a) x – 12 = 22add 12: x = 12 + 22 = 34x = 34 b) x + 19 = 229subtract 19: x = 229 – 19 = 210x = 210 c) 3x = 45divide by 3: x = 45 ÷ 3 = 15x = 15 d) multiply by 5: x = 5 × 7 = 35x = 35

2) Describe the 2 operations we do to both sides to solve, then find the solution.

 a) 2x – 12 = 22add, divide x = (22 + 12) ÷ 2 = 17 b) 7x + 19 = 229subtract, divide x = (229 - 19) ÷ 7 = 30 c) 3x – 3 = 48add, divide x = (48 + 3) ÷ 3 = 17 d) ¼ x – 3 = 1add, multiply x = (1 + 3) × 4 = 16

3) Solve these equations. Show all your work.

 a) 15x – 2 = 19 – 6xtranspose: 15x + 6x = 19 + 2Now we have 21x = 21, so x = 1 b) 4t + 5 = 35 – ttranspose: 4t + t = 35 – 5Now we have 5t = 30, so t = 6 c) 13x – 20 = 40 + xtranspose: 13x – x = 40 + 20Now we have 12x = 60, so x = 5 d) 9m – 2 = 4m + 23transpose: 9m – 4m = 23 + 2Now we have 5m = 25 , so x = 5

. (all content © MathRoom Learning Service; 2004 - ).