solve the equation by substitution, 3x+4y = 10, 2x+3y=7

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!

Learn how to solve the system of linear equations 3x + 4y = 10 and 2x + 3y = 7 using the substitution method. This step-by-step guide covers key algebraic concepts, including solving one equation for a variable and substituting into the other equation. Suitable for students in Grades 8-10.

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