Solve 2x_{1} + x_{2} - 3x_{3} = - 1; x_{1} - x_{2} + x_{3} = 3; x_{1} + x_{2} - x_{3} = 2 | Equatix - Math Solver

Fractions & Decimals

Ratios & Proportions

Basic Number Theory

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

Graphing Trigonometric Functions

Limits & Continuity

Derivatives & Applications

Integrals & Applications

Series & Sequences

Multivariable Calculus

Functions & Graphs

Polynomial & Rational Functions

Exponential & Logarithmic Functions

Sequences & Series

Descriptive Statistics

Normal Distribution

Hypothesis Testing

Regression Analysis

Linear Programming

Sets & Counting

Eigenvalues & Eigenvectors

Linear Transformations

Atomic Structure

Thermochemistry

Example

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

a⋅(b-2)=3b

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

a⋅(b-2)=3b

Question

$$\begin{cases}2x_{1}+x_{2}-3x_{3}=-1\\ x_{1}-x_{2}+x_{3}=3\\ x_{1}+x_{2}-x_{3}=2\end{cases}$$

Solve for x_1, x_2, x_3

$x_{1} = \frac{5}{2} = 2\frac{1}{2} = 2.5$
$x_{2} = \frac{9}{4} = 2\frac{1}{4} = 2.25$
$x_{3} = \frac{11}{4} = 2\frac{3}{4} = 2.75$

Short Steps Using Substitution

Solve $2x_{1}+x_{2}-3x_{3}=-1$ for $x_{2}$.

$$x_{2}=-2x_{1}+3x_{3}-1$$

Substitute $-2x_{1}+3x_{3}-1$ for $x_{2}$ in the second and third equation.

$$x_{1}-\left(-2x_{1}+3x_{3}-1\right)+x_{3}=3$$ $$x_{1}-2x_{1}+3x_{3}-1-x_{3}=2$$

Solve these equations for $x_{1}$ and $x_{3}$ respectively.

$$x_{1}=\frac{2}{3}+\frac{2}{3}x_{3}$$ $$x_{3}=\frac{3}{2}+\frac{1}{2}x_{1}$$

Substitute $\frac{2}{3}+\frac{2}{3}x_{3}$ for $x_{1}$ in the equation $x_{3}=\frac{3}{2}+\frac{1}{2}x_{1}$.

$$x_{3}=\frac{3}{2}+\frac{1}{2}\left(\frac{2}{3}+\frac{2}{3}x_{3}\right)$$

Solve $x_{3}=\frac{3}{2}+\frac{1}{2}\left(\frac{2}{3}+\frac{2}{3}x_{3}\right)$ for $x_{3}$.

$$x_{3}=\frac{11}{4}$$

Substitute $\frac{11}{4}$ for $x_{3}$ in the equation $x_{1}=\frac{2}{3}+\frac{2}{3}x_{3}$.

$$x_{1}=\frac{2}{3}+\frac{2}{3}\times \frac{11}{4}$$

Calculate $x_{1}$ from $x_{1}=\frac{2}{3}+\frac{2}{3}\times \frac{11}{4}$.

$$x_{1}=\frac{5}{2}$$

Substitute $\frac{5}{2}$ for $x_{1}$ and $\frac{11}{4}$ for $x_{3}$ in the equation $x_{2}=-2x_{1}+3x_{3}-1$.

$$x_{2}=-2\times \frac{5}{2}+3\times \frac{11}{4}-1$$

Calculate $x_{2}$ from $x_{2}=-2\times \frac{5}{2}+3\times \frac{11}{4}-1$.

$$x_{2}=\frac{9}{4}$$

The system is now solved.

$$x_{1}=\frac{5}{2}$$ $$x_{2}=\frac{9}{4}$$ $$x_{3}=\frac{11}{4}$$