Solve systems of linear equations algebraically

Solving a linear system algebraically involves using equivalent transformations, substitution, or elimination to determine the ordered pair that satisfies both linear equations, with coefficients and constants commonly expressed as integers, fractions, or decimals. The reasoning distinguishes a unique solution, no solution, and infinitely many solutions from the resulting equations and includes verifying solutions in both original equations; the scope excludes nonlinear systems, matrix-based methods, and general abstract $n$ -variable theory.

Detailed Explanation: Solve systems of linear equations algebraically

To solve a system of linear equations algebraically, find the values of $x$ and $y$ that make both equations true. One useful method is elimination, which combines the equations to eliminate one variable.

Consider the system

\begin{cases} 2x+3y=13\\ 4x-y=5 \end{cases}

Step 1: Eliminate one variable

The second equation has $-y$ , while the first has $+3y$ . Multiply the second equation by $3$ :

3(4x-y)=3(5)

This gives

12x-3y=15

The system is now

\begin{cases} 2x+3y=13\\ 12x-3y=15 \end{cases}

Step 2: Add the equations

Adding the equations eliminates $y$ :

\begin{aligned} 2x+3y&=13\\ 12x-3y&=15\\ \hline 14x&=28 \end{aligned}

Solve for $x$ :

x=2

Step 3: Substitute to find the other variable

Substitute $x=2$ into either original equation. Using $4x-y=5$ :

4(2)-y=5

8-y=5

-y=-3

y=3

Therefore, the solution is

\boxed{(2,3)}

Step 4: Verify in both original equations

Check $(2,3)$ in the first equation:

2(2)+3(3)=4+9=13

Check it in the second equation:

4(2)-3=8-3=5

Both equations are true, so $\boxed{(2,3)}$ is the solution. This system has a unique solution because the variables led to one specific ordered pair.

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Types of Solutions - Equation Pair to Zero, One, or Infinite

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