Complete the following activity to solve simultaneous equations :
`4x+ 3y = 18 " "...(1) 5x - 3y = 9 " " ` ...(2)
Adding equations (1) and (2) , `+{:(4x+3y=18" ...(1)"),(5x-3y=9" ...(2)"),(bar(squarex" "=square)):}`
` :. X = (square)/(square ) " " :. X = square`
Substituting x = 3 in equation (1) ,
` 4 xx square + 3y = 18 " " :. 3y = 18 - square " " :. 3y = square `
` :. y = 2 `
(x,y ) = `(square, square ) ` is the solution

Answer

To solve the simultaneous equations given by:

1. \( 4x + 3y = 18 \) (Equation 1)
2. \( 5x - 3y = 9 \) (Equation 2)

we will follow these steps:

### Step 1: Add the two equations
We will add Equation 1 and Equation 2 to eliminate \( y \).

\[ (4x + 3y) + (5x - 3y) = 18 + 9 \]

This simplifies to:

\[ 4x + 5x + 3y - 3y = 27 \]

Combining like terms gives:

\[ 9x = 27 \]

### Step 2: Solve for \( x \)
Now, we will solve for \( x \) by dividing both sides by 9:

\[ x = \frac{27}{9} = 3 \]

### Step 3: Substitute \( x \) back into one of the original equations
We will substitute \( x = 3 \) back into Equation 1 to find \( y \):

\[ 4(3) + 3y = 18 \]

This simplifies to:

\[ 12 + 3y = 18 \]

### Step 4: Solve for \( y \)
Now, we will isolate \( y \):

\[ 3y = 18 - 12 \]
\[ 3y = 6 \]

Dividing both sides by 3 gives:

\[ y = \frac{6}{3} = 2 \]

### Step 5: Write the solution
The solution to the simultaneous equations is:

\[ (x, y) = (3, 2) \]

### Summary
Thus, the solution to the simultaneous equations \( 4x + 3y = 18 \) and \( 5x - 3y = 9 \) is \( (3, 2) \).

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