Reading: Solving Multi-Step Equations
Learning Objective(s)
·Use properties of equality together to isolate variables and solve algebraic equations.
·Use the properties of equality and the distributive property to solve equations containing parentheses, fractions, and/or decimals.
Introduction
There are some equations that you can solve in your head quickly. For example - what is the value of y in the equation 2y = 6? Chances are you didn't need to get out a pencil and paper to calculate that y = 3. You only needed to do one thing to get the answer, divide 6 by 2.
Other equations are more complicated. Solving 
without writing anything down is difficult! That's because this equation contains not just a variable but also fractions and terms inside parentheses. This is a multi-step equation, one that takes several steps to solve. Although multi-step equations take more time and more operations, they can still be simplified and solved by applying basic algebraic rules.
Remember that you can think of an equation as a balance scale, with the goal being to rewrite the equation so that it is easier to solve but still balanced. The addition property of equality and the multiplication property of equality explain how you can keep the scale, or the equation, balanced. Whenever you perform an operation to one side of the equation, if you perform the same exact operation to the other side, you'll keep both sides of the equation equal.
If the equation is in the form, ax + b = c, where x is the variable, you can solve the equation as before. First "undo” the addition and subtraction, and then "undo” the multiplication and division.
Example | ||
Problem | Solve 3y + 2 = 11. | |
| Subtract 2 from both sides of the equation to get the term with the variable by itself. Divide both sides of the equation by 3 to get a coefficient of 1 for the variable. | |
Answer | y = 3 | |
Example | ||
Problem | Solve | |
| Add 2 from to both sides of the equation to get the term with the variable by itself. Multiply both sides of the equation by 4 to get a coefficient of 1 for the variable. | |
Answer | x = 20 | |
If the equation is not in the form, ax + b = c, you will need to perform some additional steps to get the equation in that form.
In the example below, there are several sets of like terms. You must first combine all like terms.
Example | ||
Problem | Solve 3x + 5x + 4 - x + 7 = 88. | |
| There are three like terms 3x, 5x and-x involving a variable. Combine these like terms. 4 and 7 are also like terms and can be added. The equation is now in the form ax + b = c. So, we can solve as before. Subtract 11 from both sides. Divide both sides by 7. | |
Answer | x = 11 | |
Some equations may have the variable on both sides of the equal sign. We need to "move” one of the variable terms in order to solve the equation.
Example | ||
Problem | Solve 6x + 5 = 10 + 5x. Check your solution. | |
| This equation has x terms on both the left and the right. To solve an equation like this, you must first get the variables on the same side of the equal sign. You can subtract 5x on each side of the equal sign, which gives a new equation: x + 5 = 10. This is now a one-step equation! Subtract 5 from both sides. | |
Check |
| Check your solution by substituting 5 for x in the original equation. This is a true statement, so the solution is correct. |
Answer | x = 5 | |
Here are some steps to follow when you solve multi-step equations.
Solving multi-step equations 1. If necessary, simplify the expressions on each side of the equation, including combining like terms. 2. Get all variable terms on one side and all numbers on the other side using the addition property of equality. (ax + b = c or c = ax + b) 3. Isolate the variable term using the inverse operation or additive inverse (opposite) using the addition property of equality. 4. Isolate the variable using the inverse operation or multiplicative inverse (reciprocal) using the multiplication property of equality to write the variable with a coefficient of 1. 5. Check your solution by substituting the value of the variable in the original equation. |
The examples below illustrate this sequence of steps.
Example | |||
Problem | Solve for y. -20y + 15 = 2 - 16y + 11 | ||
| Step 1. On the right side, combine like terms: 2 + 11 = 13. Step 2. Add 20y to both sides to remove the variable term from the left side of the equation. Step 3. Subtract 13 from both sides. Step 4. Divide 4y by 4 to solve for y. | ||
Check |
| Step 5. To check your answer, substitute | |
Answer |
| ||
for y in the original equation. The statement 5 = 5 is true, so y = 
Advanced Example | |||
Problem | Solve 3y + 10.5 = 6.5 + 2.5y. Check your solution. | ||
| This equation has y terms on both the left and the right. To solve an equation like this, you must first get the variables on the same side of the equal sign. | ||
| Add -2.5y to both sides so that the variable remains on one side only. | ||
| Now isolate the variable by subtracting 10.5 from both sides. | ||
| Multiply both sides by 10 so that 0.5y becomes 5y, then divide by 5. | ||
Check |
| Check your solution by substituting -8 in for y in the original equation. This is a true statement, so the solution is correct. | |
Answer | y = -8 | ||
Advanced Question Identify the step that will not lead to a correct solution to the problem.
A) Multiply both sides of the equation by 2. B) Add C) Add D) Rewrite |
to both sides of the equation.
to the left side, and add 
to the right side.
as 
.More complex multi-step equations may involve additional symbols such as parentheses. The steps above can still be used. If there are parentheses, you use the distributive property of multiplication as part of Step 1 to simplify the expression. Then you solve as before.
The Distributive Property of Multiplication |
For all real numbers a, b, and c, a(b + c) = ab + ac. |
What this means is that when a number multiplies an expression inside parentheses, you can distribute the multiplication to each term of the expression individually. Then, you can follow the routine steps described above to isolate the variable to solve the equation.
Example | |||
Problem | Solve for a. 4(2a + 3) = −3(a − 1) + 31 | ||
| Apply the distributive property to expand4(2a + 3) to 8a + 12 and −3(a - 1) to−3a + 3. Combine like terms. Add 3a to both sides to move the variable terms to one side. Subtract 12 to isolate the variable term. Divide both terms by 11 to get a coefficient of 1. | ||
Answer | a = 2 | ||
In which of the following equations is the distributive property properly applied to the equation 2(y +3) = 7? A) y + 6 = 7 B) 2y + 6 = 14 C) 2y + 6 = 7 D) 2y + 3 = 7 |
If you prefer not working with fractions, you can use the multiplication property of equality to multiply both sides of the equation by a common denominator of all of the fractions in the equation. See the example below.
Example | ||
Problem | Solve | |
| Multiply both sides of the equation by 4, the common denominator of the fractional coefficients. Use the distributive property to expand the expressions on both sides. Multiply. Add 3x to both sides to move the variable terms to only one side. Add 12 to both sides to move theconstant terms to the other side. Divide to isolate the variable. | |
Answer |
| |
Of course, if you like to work with fractions, you can just apply your knowledge of operations with fractions and solve.
Example | ||||
Problem | Solve | |||
| Add
Add 3 to both sides to get the constant terms on the other side. To get a coefficient of 1, multiply the variable term by its multiplicative inverse. | |||
Answer |
| |||
Advanced Example | ||
Problem | Solve | |
| Solving this equation will require multiple steps. Begin by evaluating 32 = 9. | |
| Now distribute the | |
| Multiply both sides of the equation by 18, the common denominator of the fractions in the problem. Use the distributive property to expand the expression on the left side. Then remove a factor of 1 from both sides. On the left, you can think of | |
| Continue solving for a using the distributive property. Then isolate the variable, and solve the remaining one-step problem. | |
Check
| Check your solution by substituting This is a true statement, so the solution is correct. | |
Answer |
| |
. On the right, you can think of 
.
in for a in the original equation.
To clear the fractions from 3 6 9 27 A) 9 B) 9 or 27 C) 6 D) 3 or 9 |
, we can multiply both sides of the equation by which of the following numbers?Regardless of which method you use to solve equations containing variables, you will get the same answer. You can choose the method you find easier! Remember to check your answer by substituting your solution into the original equation.
Just as you can clear fractions from an equation, you can clear decimals from the equation in the same way. Find a common denominator and use the multiplication property of equality to multiply both sides of the equation.
Example | ||
Problem | Solve 0.4x - 0.25 = 1.75 by clearing the decimals first. | |
| 0.4 ( have a common denominator of 100. Multiply both sides by 100. Apply the distributive property to clear the parentheses. Solve as before. Add 25 to both sides. Divide both sides by 40. | |
Check: |
| Substitute x = 5 into the original equation. Evaluate. The solution checks. |
Answer |
| |
) and 0.25 (
) and 1.75 (
)
Advanced Question Solve for a: A) a = 2 B) a = 1 C) a = 0 D) a = -2 |
Summary
Complex, multi-step equations often require multi-step solutions. Before you can begin to isolate a variable, you may need to simplify the equation first. This may mean using the distributive property to remove parentheses, or multiplying both sides of an equation by a common denominator to get rid of fractions. Sometimes it requires both techniques.
Permissions
This reading is taken from the Developmental Math Open Program created by The NROC Project. It is available under a Creative Commons license.
Последнее изменение: четверг, 25 августа 2016, 09:19