How to solve 3x + 1 < 16

Q: How to solve 3x + 1 < 16

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve 3x + 1 < 16 | Equatix

Integers & Absolute Values

Factors & Multiples

Prime Factorization

Basic Equations & Inequalities

Coordinate Plane & Graphing

Linear Equations & Inequalities

Quadratic Equations

Polynomials & Factoring

Functions & Graphs

Systems of Equations

Trigonometric Functions

Trigonometric Identities & Equations

Right Triangle Trigonometry

Unit Circle

Graphing Trigonometric Functions

Limits & Continuity

Derivatives & Applications

Integrals & Applications

Series & Sequences

Multivariable Calculus

Functions & Graphs

Polynomial & Rational Functions

Exponential & Logarithmic Functions

Conic Sections

Sequences & Series

Descriptive Statistics

Eigenvalues & Eigenvectors

Linear Transformations

Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$3x+1<16$$

Solve for x

$x<5$

Solution Steps

Subtract $1$ from both sides.

$$3x<16-1$$

Subtract $1$ from $16$ to get $15$.

$$3x<15$$

Divide both sides by $3$. Since $3$ is positive, the inequality direction remains the same.

$$x<\frac{15}{3}$$

Divide $15$ by $3$ to get $5$.

$$x<5$$

Show Solution

Basic Math

Pre-Algebra

Algebra

Trigonometry

Calculus

Precalculus

Statistics

Finite Math

Linear Algebra

Chemistry

Example

Question

Solve for x

Solution Steps