Solve the following pair of equations to find an expression for x in terms of t. 2x + 3y = 10 9y - 4t = 15

To solve the given pair of equations for \( x \) in terms of \( t \), we can proceed with the following steps:

### Given Equations:
1. \( 2x + 3y = 10 \)  \(\quad\) \(\text{(Equation 1)}\)
2. \( 9y - 4t = 15 \)  \(\quad\) \(\text{(Equation 2)}\)

### Step 1: Solve Equation 2 for \( y \) in terms of \( t \)

Start with Equation 2:

\[
9y - 4t = 15
\]

Isolate \( y \):

\[
9y = 15 + 4t
\]

\[
y = \frac{15 + 4t}{9}
\]

### Step 2: Substitute \( y \) in Equation 1

Now, substitute the expression for \( y \) into Equation 1:

\[
2x + 3\left(\frac{15 + 4t}{9}\right) = 10
\]

### Step 3: Solve for \( x \) in terms of \( t \)

First, simplify the equation:

\[
2x + \frac{3(15 + 4t)}{9} = 10
\]

\[
2x + \frac{45 + 12t}{9} = 10
\]

Multiply through by 9 to eliminate the fraction:

\[
18x + 45 + 12t = 90
\]

Simplify by subtracting 45 from both sides:

\[
18x + 12t = 45
\]

Finally, solve for \( x \):

\[
18x = 45 - 12t
\]

\[
x = \frac{45 - 12t}{18}
\]

Simplify the expression:

\[
x = \frac{15 - 4t}{6}
\]

### Final Answer:

\[
x = \frac{15 - 4t}{6}
\]

Would you like further details or have any questions?

Here are five related questions for further practice:

1. How would you solve for \( y \) in terms of \( t \) from the given equations?
2. What are the steps to eliminate a variable when solving a system of linear equations?
3. Can you express \( t \) in terms of \( x \) using the final expression for \( x \)?
4. How would the solution change if one equation was non-linear?
5. How can substitution be used to solve systems with three variables?

**Tip:** Always simplify the expression as much as possible to make the solution more interpretable.

Solve the following pair of equations to find an expression for x in terms of t. 2x + 3y = 10, 9y - 4t = 15

Learn how to solve the system of linear equations 2x + 3y = 10 and 9y - 4t = 15 to express x in terms of t. This step-by-step guide covers key algebraic concepts like the substitution method and solving linear equations. Suitable for students in Grades 9-12.

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