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Roots - Radicals 1
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Sum of the Roots of a Quadratic
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Solving Equations 3
Solving Quadratic Equations
Parent and Family Graphs
Collecting Like Terms
nth Roots
Power of a Quotient Property of Exponents
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The Quadratic Formula
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Polar Form of a Complex Number
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Algebra
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Higher Degrees and Variable Exponents
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Writing a Rational Expression in Lowest Terms
Solving Quadratic Inequalities with a Sign Graph
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The Square of a Binomial
Properties of Negative Exponents
Inverse Functions
fractions
Rotating an Ellipse
Multiplying Numbers
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Solving Equations with One Log Term
Combining Operations
The Ellipse
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Graphing Inequalities in Two Variables
Solving Trigonometric Equations
Adding and Subtracting Fractions
Simple Trinomials as Products of Binomials
Ratios and Proportions
Solving Equations
Multiplying and Dividing Fractions 2
Rational Numbers
Difference of Two Squares
Factoring Polynomials by Grouping
Solving Equations That Contain Rational Expressions
Solving Quadratic Equations
Dividing and Subtracting Rational Expressions
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Order of Operations
Solving Nonlinear Equations by Substitution
The Distance and Midpoint Formulas
Linear Equations
Graphing Using x- and y- Intercepts
Properties of Exponents
Solving Quadratic Equations
Solving One-Step Equations Using Algebra
Relatively Prime Numbers
Solving a Quadratic Inequality with Two Solutions
Quadratics
Operations on Radicals
Factoring a Difference of Two Squares
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Solving Quadratic Equations by Factoring
Graphing Logarithmic Functions
Simplifying Expressions Involving Variables
Adding Integers
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Factoring Completely General Quadratic Trinomials
Using Patterns to Multiply Two Binomials
Adding and Subtracting Rational Expressions With Unlike Denominators
Rational Exponents
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Solving Equations

Most of algebra involves either simplifying expressions (by writing equivalent expressions) or solving equations (by writing equivalent equations). We can use the addition and multiplication principles to produce equivalent equations, like x = 5 from which the solution - in this case, 5- is obvious. Here we will find that more complicated equations can be solved by using both principles together and by using the commutative, associative, and distributive laws to write equivalent expressions. An important strategy for solving a new look like a problem we already know how to solve. This is precisely the approach taken in this section.

What is new in this section appears in the early steps of each example. Without a solid grasp of how and when to use the addition and multiplication principles, the problems in this section will seem much more difficult than they really are.

Using both the Addition and Multiplication Principles

In the expression 5 + 3x the variable x is multiplied by 3 and then 5 is added. To reverse these steps, we first subtract 5 and then divide by 3. Thus, to solve 5 + 3x = 17 we first subtract 5 from each side and then divide both sides by 3.

Example 1

Solve: 5 + 3x = 17

Solution

We have

5 + 3x = 17

5 + 3x - 5 = 17 - 5 Using the addition principle: subtracting 5 from both sides (adding -5)

5 + (- 5) + 3x = 12 Using a commutative law. Try to perform this step mentally.

First isolate the x-term

3x = 12 Simplifying

Using the multiplication principle: dividing both sides by 3 multiplying by

Then isolate x

x = 4 Simplifying

Check:

We use the rules for order of operations: Find the product , and then add.

The solution is 4.

Multiplication by a negative number and subtraction are handled in much the same way.

Example 2

Solve -5x -6 = 16.

Solution

In -5x -6 we multiply first and then subtract. To reverse these steps, we first add 6 and then divide by -5.

-5x -6 = 16

-5x -6 + 6 = 16 + 6 Adding 6 to both sides.

Dividing both sides by -5

Simplifying

Check:

The soultion is .

Example 3

Solve 45 - t = 13.

Solution

45 - t = 13

45 - t - 45 = 13 - 45 Substracting 45 from both sides.

Try to do these steps mentally

-t = -32 Try to go directly to this step

(-1)(-t) = (-1)(-32) Multiply both sides by -1 (Dividing by -1 would also work)

t = 32

Check:

The soultion is 32.

As our skills improve, many of the steps can be streamlined.

Example 4

Solve 16.3 - 7.2y = -8.18

Solution

We have

16.3 - 7.2y = -8.18

- 7.2y = -8.18 -16.3 Subtracting 16.3 from both sides. We write the subtraction of 16.3 on the right side and remove 16.3 from the left side.

- 7.2y = -24.48

Dividing both sides by -7.2.We write the division by -7.2 on the right side and remove the from the left side.

y = 3.4

Check:

The solution is 3.4.

Combining Like Terms

If like terms appear on the same side of an equation, we combine them and then solve. Should like terms appear on both sides of an equation, we can use the addition principle to rewrite all like terms on one side

Example 5

Solve:

a) 3x + 4x = -14

b) 2x - 4 = -3x + 1

c) 6x + 5 -7x = 10 - 4x + 7

d) 2 - 5(x + 5) = 3(x - 2) - 1

Solution

a) 3x + 4x = -14

7x = -14 Combining like terms

Dividing both sides by 7

x = -2

The check is left to the student. The solution is -2.

To solve 2x - 4 = -3x + 1 we must first write only variable terms on one side and only constant terms on the other. This can be done by adding 4 to both sides, to get all constant terms on the right, and 3x to both sides, to get all variable terms on the left. We can add 4 first, or 3x first, or do both in one step.

2x - 4 = -3x + 1

2x - 4 + 4 = -3x + 1 + 4 Adding 4 to both sides

2x = -3x + 5 Simplifying

2x + 3x = -3x + 3x + 5 Adding 3x to both sides

5x = 5 Combining like terms and simplifying

Dividing both sides by 5

x = 1 Simplifying

Check:

The solution is 1.

c) 6x + 5 - 7x = 10 - 4x + 7

-x + 5 = 17 - 4x Combining like terms on both sides

-x + 5 + 4x = 17 - 4x + 4x Adding 4x to both sides

5 + 3x = 17 Simplifying. This is identical to Example 1

3x = 12 Subtracting 5 from both sides and simplifying

x = 4 Dividing both sides by 3 and simplifying.

Check:

The solution is 4.

d) 2 - 5(x + 5) = 3(x - 2) - 1

2 - 5x - 25 = 3x - 6 - 1 Using the distributive law. This is now similar to part (c) above.

-5x - 23 = 3x - 6 Combining like terms on both sides

Adding 7 and 5x to both sides. This isolates the x-Simplifying

-16 = 8x Simplifying

-2 = x Dividing both sides by 8

The student can confirm that -2 checks and is the solution.

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