Sensor de Temperatura

Algebraic equations often include an expression in parentheses which is multiplied by a term. In those situations we must use the distributive property to remove the parentheses before proceeding with solving.

4

7

=

𝑥

10

Remember: The distributive property of multiplication

over addition can be represented as a(b + c) = ab + ac. The formula also applies to subtraction: a(b – c) = ab – ac.

In the first book we noted that the distributive property didn’t seem all that useful in situations where it was easy to just add or subtract whatever was in parentheses. In algebra, though we will often work with expressions such as 3(x + 2) where it isn’t possible to combine what is in parentheses. The only way to remove the parentheses is if we distribute the term that is outside.

Let’s look at the equation 3(x + 2) = -15. The truth is that there are two different ways to solve this equation, one of which is to start by dividing both sides by 3. While that will work in equations such as this, it will not be that simple in other equations which look quite similar.

3(x + 2) = -15

3x + 2·3 = -15

3x + 6 = -15

3x = -21

x = -7

Check:

3(x + 2) = -15

3(-7 + 2) = -15

3(-5) = -15

-15 = -15

Using the distributive property of multiplication over addition to remove the parentheses in an algebraic equation

M O R E C O M P L I C A T E D A L G E B R A I C E Q U A T I O N S

The best way to proceed is to first use the use the distri- butive property to remove the parentheses. Once that is done, we will be able to proceed just like we did in all of our previous examples.

Notice how we started by distributing the 3 over the x and the 2 in the parentheses. Once that was done we were left with a basic two-step equation no different than we saw in previous examples.

Remember: When an algebraic equation includes a term

which multiplies an expression in parentheses, it is usually best to first use the distributive property to remove the parentheses before proceeding.

Let’s use the distributive property to remove the paren- theses in the expression -5(x – 7). The recommended method is to follow the pattern of a(b – c) = ab – ac. In this case, our a term is -5, our b term is x, and our c term is 7. The problem is illustrated at left. Notice how the -5 first multiplies the x, and then it multiplies the 7. Those two products are connected by subtraction. The hardest part of handling a problem like this is remembering all the rules for signed number arithmetic. If necessary, review the first book for details.

-5(x – 7)

-5x – (-5 · 7)

-5x – (-35)

-5x + 35

Understand that while it is possible to treat the minus sign before the 7 as a negative sign for the 7, if we do, we will no longer have an operation connecting the x and the -7. We are allowed to put in an implied addition sign between them, but all of that is unnecessary work if you proceed as described above.

PRACTICE EXERCISES AND REVIEW

This chapter covered quite a bit of material. Remember that before solving an algebraic equation, you should check to see if any combining of like terms can be done. You must handle each side of the equation separately when combining.

No matter how complicated an equation is, all of the constants will have to be “moved over” and combined on one side, and all of the variable terms will have to be “moved over” and combined on the other side. It does not matter which side ends up with which. You will get the same answer in either case.

An example of a common equation form is 3x + 5 = 17. The best way to proceed is to first subtract 5 from each side, giving us 3x = 12. Then divide both sides by 3 to get x = 4. Be very careful with your signed number arithmet- ic in any equation which involves negatives.

M O R E C O M P L I C A T E D A L G E B R A I C E Q U A T I O N S

Review the procedure for solving a mean (average) problem involving an unknown value. The process is always the same.

Remember: In equations which involve a variable times

a fraction, the best way to proceed is to multiply both sides by the fraction’s reciprocal.

Remember: In the simple interest formula, time must

stated be in terms of years. That means that 6 months must be represented in the formula as 0.5. Be careful to determine if a problem is asking about just the interest, or the interest with the principal added back in. Remem- ber to convert the interest rate percent into a decimal.

Remember: In a proportion problem involving a varia-

ble, set the cross products equal to each other, and solve the resulting one-step equation.

Remember: The distributive property can be used to

eliminate parentheses in expressions of the form a(b + c), even in expressions that involve variables. When distri- buting a negative value, be certain that you are properly adhering to the rules of signed number multiplication. Also remember that an expression such as x – 5 could be rewritten and treated as the equivalent x + (-5).

For practice, try these exercises:

1) Solve for x: −8 − −7 + 4 = 7𝑥 − 9𝑥 + 12𝑥 2) Solve for x: −2𝑥 − 9 = 15

3) Solve for x: −9𝑥 − 3 = −13𝑥 + 5

4) A student received grades of 82, 96, 81, 42, and 63. What grade must s/he get on the next test to have a mean of 75?

5) A person invests $3600 for 6 months in an account which offers an APY of 2.50%. How much money in total will s/he have at the end of the time period? 6) A person wants to earn $475 on a $4100 investment.

If s/he is willing to invest the money for 3 years, what is the minimum interest rate s/he must find? 7) Solve for x: ₁₁3 =13_𝑥

8) Solve for x: 5(x + 6) = 42 9) Solve for x: -10(x – 4) = -15

SO NOW WHAT?

Before progressing to the next chapter, it is essential that you fully understand all of the concepts in this one. If you do not, you will have tremendous difficulty with later work, and will need to return to this chapter.