Find the unknown number. If half of a number is added to 2 5 \frac25 , the result is the same as subtracting 1 10 \frac{1}{10} from the number.

Solve this problem, we're going to use our five steps for solving word problems. Where our first step is to understand the problem by reading the problem, drawing a picture when needed, and identifying and naming our unknown. So let's read this problem. If half of a number is added to two fifths, the result is the same as subtracting one tenth from that number. Drawing a picture isn't really going to help us here. So let's go ahead and identify name our unknowns. We have half of a number, so that's our unknown, and we'll call it x. Subtracting one tenth from the number, so that's the same number. So we're going to also call that x. So next, let's build our equation. And in order to do this, we're going to use some translation techniques that we've learned previously. So half of a number. Well, if our number is x, half of that number is x divided by two. So half of a number is added to two fifths. So let's take that and add two fifths. The result is the same. That's another way for saying equal to. The result is the same means equal to as subtracting one tenth from that number. So we have our number x and we're going to subtract one tenth from it. So now that we have our equation, we can solve it. Here we're going to be solving for x. So my first step in trying to isolate x is I'm going to go ahead and add one tenth to both sides. And in order to combine these two fractions, we need to have common denominators. So I'm going to multiply two fifths by two. So that way they both have a denominator of 10. So I have x over two plus two fifths times two is four tenths plus one tenth equals x. And then we're going to combine our like terms here and move our x over two to the other side by subtracting it. Simplifying, we get four tenths plus one tenth is five tenths, and 1x minus one half x is one half x. So further solving for x, we're going to multiply both sides by two, and we get 10 over 10 equals x. And 10 over 10 simply reduces to one, so we have x equals one. Last, let's check our answer. We can do this by taking x equals one and plugging it back into the equation that we built at the beginning. So one over two plus two fifths equals one minus one tenth. In order to add these two fractions, we need to get common denominators again. So we're going to multiply this one by five and multiply this one by two, and we get five tenths plus four tenths. And this is equal to one minuteus one tenth. So five tenths plus four tenths is nine tenths, and one minus one tenth is also nine tenths. So we can feel confident that our answer here is correct.

Indice

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1. Review of Real Numbers2h 43m

Simplifying Fractions
30m

Evaluating Exponents
29m

Evaluating Expressions
15m

Translating Phrases to Expressions
16m

Translating Sentences to Equations
9m

The Distributive Property
17m

Simplifying Expressions
44m

2. Linear Equations and Inequalities5h 35m

Solving Linear Equations
2h 23m

Linear Inequalities in One Variable
46m

Set Operations and Compound Inequalities
1h 7m

Absolute Value Equations
38m

Absolute Value Inequalities
39m

3. Solving Word Problems2h 46m

Introduction to Problem Solving
37m

Formulas
25m

Percent Problem Solving
59m

Mixture Problem Solving
43m

4. Graphs and Functions5h 12m

The Rectangular Coordinate System
44m

Graph Linear Equations in Two Variables
24m

Graph Linear Equations Using Intercepts
23m

Slope of a Line
44m

Slope-Intercept Form
38m

Point Slope Form
22m

Linear Inequalities in Two Variables
28m

Introduction to Relations and Functions
53m

Function Notation
15m

Composition of Functions
17m

5. Systems of Linear Equations1h 53m

Solving Systems of Linear Equations by Graphing
29m

Solving Systems of Linear Equations by Substitution
15m

Solving Systems of Linear Equations by Elimination
45m

Systems of Linear Inequalities
23m

6. Exponents, Polynomials, and Polynomial Functions3h 17m

The Product Rule
9m

The Quotient Rule
10m

Intro to the Power Rules
17m

The Power of a Quotient Rule
13m

Negative Exponents
24m

Simplifying Exponential Expressions Using All Exponent Rules
6m

Intro to Polynomials
21m

Adding and Subtracting Polynomials
20m

Multiplying Polynomials
38m

Special Products
34m

7. Factoring2h 49m

Factoring by Greatest Common Factor & Grouping
38m

Factoring Trinomials of the Form x² + bx + c
11m

Factoring Trinomials of the Form ax² + bx + c
37m

Factoring Special Products
41m

Quadratic Equations & Applications
41m

8. Rational Expressions and Functions3h 44m

Simplifying Rational Expressions
42m

Multiplying and Dividing Rational Expressions
25m

Adding and Subtracting Rational Expressions with Common Denominators
19m

Least Common Denominators
32m

Adding and Subtracting Rational Expressions with Different Denominators
32m

Rational Equations
44m

Direct & Inverse Variation
27m

9. Roots, Radicals, and Complex Numbers3h 57m

Introduction to Radical Expressions
47m

Simplifying Radical Expressions
47m

Adding and Subtracting Radicals
19m

Multiplying, Dividing, and Rationalizing Radicals
53m

Introduction to Complex Numbers
18m

Multiplying and Dividing Complex Numbers
51m

10. Quadratic Equations and Functions3h 1m

The Square Root Property
25m

Completing the Square
26m

The Quadratic Formula
40m

Graphing Quadratic Equations
1h 28m

11. Inverse, Exponential, & Logarithmic Functions2h 30m

Introduction to Inverse Functions
25m

Exponential Functions
35m

Introduction to Logarithmic Functions
33m

Properties of Logarithms
54m

12. Conic Sections & Systems of Nonlinear Equations2h 24m

Graphing Circles
32m

Ellipses
35m

Parabolas
35m

Hyperbolas
41m

13. Sequences, Series, and the Binomial Theorem1h 50m

Introduction to Sequences
40m

Arithmetic Sequences
27m

Geometric Sequences
29m

Factorials
13m

3. Solving Word Problems

Introduction to Problem Solving

Intermediate Algebra

3. Solving Word Problems

Introduction to Problem Solving

Problema precedenteProssimo problema

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Find the unknown number.
If half of a number is added to $\frac{2}{5}$ , the result is the same as subtracting $\frac{1}{10}$ from the number.

$-$\frac$15$

$-1$

$1$

$-$\frac$35$

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Problema precedenteProssimo problema

Guida verificata passo dopo passo

Let the unknown number be represented by the variable $x$.

Translate the problem statement into an equation: "If half of a number is added to two fifths, the result is the same as subtracting one tenth from the number." This can be written as: $\frac{1}{2}x + \frac{2}{5} = x - \frac{1}{10}$.

To solve for $x$, first eliminate the fractions by finding the least common denominator (LCD) of 2, 5, and 10, which is 10. Multiply every term in the equation by 10 to clear the denominators.

After clearing denominators, simplify the resulting equation and collect like terms on each side to isolate the variable $x$.

Finally, solve the simplified linear equation for $x$ by performing the necessary algebraic operations such as addition, subtraction, and division.

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