Problem 88
Question
Simplify each expression. \(\frac{1}{4}(8 x+16)-\frac{1}{5}(20 x-15)\)
Step-by-Step Solution
Verified Answer
-2x + 7
1Step 1 - Distribute the fractions
Distribute the \(\frac{1}{4}\) to both terms inside the parentheses, and distribute \(\frac{1}{5}\) to both terms inside the second parentheses. This gives:\[\frac{1}{4}(8x + 16) - \frac{1}{5}(20x - 15) = \frac{1}{4} \times 8x + \frac{1}{4} \times 16 - \frac{1}{5} \times 20x + \frac{1}{5} \times 15\]
2Step 2 - Simplify each term
Calculate each multiplication:\[\frac{1}{4} \times 8x = 2x, \frac{1}{4} \times 16 = 4, \frac{1}{5} \times 20x = 4x, \frac{1}{5} \times 15 = 3\]So the expression becomes:\[2x + 4 - 4x + 3\]
3Step 3 - Combine like terms
Combine the like terms for \(x\) and constants:\[2x - 4x + 4 + 3 = -2x + 7\]
Key Concepts
distributive propertylike termscombining constants
distributive property
The distributive property is a key concept in algebra for simplifying expressions. It states that for any numbers or variables, you can distribute a factor across terms inside parentheses. For example, if you have \[a(b + c) = ab + ac\]. This means you multiply each term inside the parentheses by the factor outside.
In our exercise, we applied the distributive property to \[\frac{1}{4}(8x + 16)\] and \[\frac{1}{5}(20x - 15)\] respectively. This involves:
Breaking it down, we get \[\frac{1}{4} \times 8x + \frac{1}{4} \times 16\] and \[\frac{1}{5} \times 20x + \frac{1}{5} \times (-15)\]. This simplifies to \[2x + 4 - 4x + 3\].
In our exercise, we applied the distributive property to \[\frac{1}{4}(8x + 16)\] and \[\frac{1}{5}(20x - 15)\] respectively. This involves:
- Multiplying \[\frac{1}{4}\] by \[8x\] and \[16\].
- Multiplying \[\frac{1}{5}\] by \[20x\] and \[15\].
Breaking it down, we get \[\frac{1}{4} \times 8x + \frac{1}{4} \times 16\] and \[\frac{1}{5} \times 20x + \frac{1}{5} \times (-15)\]. This simplifies to \[2x + 4 - 4x + 3\].
like terms
Like terms are terms that have the same variable raised to the same power. Only like terms can be combined. Identifying them is crucial for simplifying expressions.
In our expression \[2x + 4 - 4x + 3\], the like terms are \[2x\] and \[-4x\]. Both have the variable \[x\] raised to the power of one. The constants \[4\] and \[3\] are also like terms because they are just numbers without variables.
To identify like terms, follow these steps:
In our expression \[2x + 4 - 4x + 3\], the like terms are \[2x\] and \[-4x\]. Both have the variable \[x\] raised to the power of one. The constants \[4\] and \[3\] are also like terms because they are just numbers without variables.
To identify like terms, follow these steps:
- Look for terms that have the same variable and exponent.
- Constant terms are naturally like terms.
combining constants
Combining constants helps to further simplify algebraic expressions. Constants are numbers without variables. They simplify by straightforward addition or subtraction.
For instance, in \[2x + 4 - 4x + 3\], the terms \[4\] and \[3\] are constants. Adding them gives \[4 + 3 = 7\].
Add these to our combined like terms to get the final simplified expression. Our result \[ -2x + 7 \] includes the simplified constant sum.
This makes it easier to handle the expression in further mathematical operations or solve for variables.
For instance, in \[2x + 4 - 4x + 3\], the terms \[4\] and \[3\] are constants. Adding them gives \[4 + 3 = 7\].
Add these to our combined like terms to get the final simplified expression. Our result \[ -2x + 7 \] includes the simplified constant sum.
This makes it easier to handle the expression in further mathematical operations or solve for variables.
Other exercises in this chapter
Problem 88
Perform each indicated operation. \(\frac{-5(2)+[3(-2)-4]}{-3-(-1)}\)
View solution Problem 88
Perform each indicated operation. $$ |-4-2|-|-8-1| $$
View solution Problem 89
Use the distributive property to rewrite each expression. $$ -3(8 x+3 y+4 z) $$
View solution Problem 89
Perform each indicated operation. $$ \left(-\frac{3}{8}-\frac{2}{3}\right)-\left(-\frac{9}{8}-3\right) $$
View solution