We are given the system of equations:

1. $3x + 4y = 10$
2. $2x + 3y = 7$

We'll solve this system using the substitution method by expressing one variable in terms of the other and substituting it into the second equation.

Step 1: Solve one equation for one variable.

Let's solve the first equation for $x$:

$3x + 4y = 10$

Isolate $x$:

$3x = 10 - 4y$

$x = \frac{10 - 4y}{3}$

Step 2: Substitute this expression for $x$ into the second equation.

Now substitute $x = \frac{10 - 4y}{3}$ into the second equation:

$2x + 3y = 7$

Substitute:

$2 \left( \frac{10 - 4y}{3} \right) + 3y = 7$

Simplify:

$\frac{2(10 - 4y)}{3} + 3y = 7$

$\frac{20 - 8y}{3} + 3y = 7$

Multiply the entire equation by 3 to eliminate the fraction:

$20 - 8y + 9y = 21$

Simplify:

$20 + y = 21$

Solve for $y$:

$y = 21 - 20$

$y = 1$

Step 3: Substitute $y = 1$ back into the expression for $x$.

Substitute $y = 1$ into the equation $x = \frac{10 - 4y}{3}$:

$x = \frac{10 - 4(1)}{3}$

$x = \frac{10 - 4}{3}$

$x = \frac{6}{3} = 2$

Final Answer:

The solution to the system of equations is:

$x = 2, \quad y = 1$

Would you like more details or have any questions? Here are some related problems:

1. Solve using substitution: $x + 2y = 5$, $3x - y = 2$
2. Solve by elimination: $5x + 2y = 12$, $7x - 4y = 9$
3. Solve the system $4x + 5y = 23$, $3x + y = 9$
4. Find the intersection point of $y = 2x + 3$ and $y = -x + 1$
5. Solve $6x + 4y = 14$ and $x + 2y = 4$ by substitution.

Tip: Always simplify one equation for one variable before substitution for an easier process!