Example: Solve the system

$2x + 3y = 10 \quad \text{(Equation A)}$ $5x - 4y = 7 \quad \text{(Equation B)}$

Solving by Elimination

1. Eliminate one variable:

• Multiply Equation $A$ by $5$: $10x + 15y = 50$
• Multiply Equation $B$ by $2$: $10x - 8y = 14$
• Subtract the new equations: $(10x + 15y) - (10x - 8y) = 50 - 14$

$23y = 36$

$\therefore \boxed {y = \frac{36}{23}}$

2. Substitute y back into one of the original equations to find $x$:

• Substitute into Equation $A$: $2x + 3\left(\frac{36}{23}\right) = 10$ $2x + \frac{108}{23} = 10$ $2x = 10 - \frac{108}{23}$ $2x = \frac{230 - 108}{23}$ $2x = \frac{122}{23}$ $x = \frac{61}{23}$

Using a Calculator

These equations can also be easily solved using a calculator if in the form:

$ax + by = k$

$cx + dy = m$

where $a, b, c, d, k, m \in \mathbb{R}$.

1. Select the "Simultaneous Equation" mode.
2. Input the coefficients for the equations.
3. The calculator will display the solutions for $x$ and $y$:

• $x = \frac{61}{23}$
• $y = \frac{36}{23}$

Step 1: Make one variable's coefficients equal

We want to eliminate one variable. To do that, we can make the coefficients of one variable the same in both equations.

Let's eliminate $y$. In equation (1), the coefficient of $y$ is 1, and in equation (2), it's $-2$. To match these, we can multiply equation (1) by 2, so that both equations have a coefficient of $y$ involving 2.

Multiply equation (1) by 2:

This gives:

Step 2: Add or subtract the equations

Now, we will add equations (3) and (2) to eliminate $y$.

Simplifying this:

Step 3: Solve for $x$

Now, solve for $x$ by dividing both sides by 7:

So:

Step 4: Substitute $x$ back into one of the original equations

Now, substitute $x = \frac{18}{7}$ into one of the original equations to solve for $y$. Let's use equation (1):

Substitute $x = \frac{18}{7}$:

This simplifies to:

Now, subtract $\frac{36}{7}$ from both sides:

Convert 7 into a fraction:

Now simplify:

So:

Final Answer:

The solution to the system of equations is:

Or approximately: