A computer dealer sells two models of computer A and B. In one particular week he sells 6 model A’s and 9 model B’s. The profit on these sales is P 585. He would make the same profit from selling 3 model of A’s and 2 model B’s. If the profit on a model A computer is P x and on a model B computer is P y,

a)	 Obtain two simultaneous equations in x and y to express these facts.	[4 marks]

b)	Solve these equations for x and y.						[4 marks]

Let's break this down step by step:

### Part (a) - Obtain two simultaneous equations in \(x\) and \(y\)

Let:
- \(x\) be the profit from selling one model A computer.
- \(y\) be the profit from selling one model B computer.

We are given two situations:
1. The dealer sells 6 model A’s and 9 model B’s, and the total profit is P 585.
   - This gives the equation:
   \[
   6x + 9y = 585
   \]

2. The dealer would make the same profit from selling 3 model A’s and 2 model B’s.
   - This gives the equation:
   \[
   3x + 2y = 585
   \]

Thus, the two simultaneous equations are:
\[
6x + 9y = 585 \tag{1}
\]
\[
3x + 2y = 585 \tag{2}
\]

### Part (b) - Solve these equations for \(x\) and \(y\)

Now, we solve the system of equations:

#### Step 1: Simplify Equation (1)
Divide the first equation by 3 to make it easier to work with:
\[
2x + 3y = 195 \tag{3}
\]
So now we have two equations:
\[
2x + 3y = 195 \tag{3}
\]
\[
3x + 2y = 585 \tag{2}
\]

#### Step 2: Eliminate one variable
We can eliminate one variable by multiplying the equations to make the coefficients of \(x\) or \(y\) the same.

Multiply equation (3) by 3:
\[
6x + 9y = 585 \tag{4}
\]

Multiply equation (2) by 2:
\[
6x + 4y = 1170 \tag{5}
\]

#### Step 3: Subtract the equations
Now subtract equation (4) from equation (5):
\[
(6x + 4y) - (6x + 9y) = 1170 - 585
\]
\[
6x + 4y - 6x - 9y = 585
\]
\[
-5y = 585
\]
\[
y = -117
\]

#### Step 4: Solve for \(x\)
Substitute \(y = -117\) into one of the original equations, say equation (3):
\[
2x + 3(-117) = 195
\]
\[
2x - 351 = 195
\]
\[
2x = 195 + 351
\]
\[
2x = 546
\]
\[
x = 273
\]

### Final Answer:
The profit per model A computer is \(x = 273\), and the profit per model B computer is \(y = -117\).

---
Do you have any questions, or would you like further details on any of these steps? 

Here are 5 related questions to further explore:
1. How can we verify the solutions for \(x\) and \(y\) using substitution?
2. What if the profit equation for model B changes to a different value?
3. How would the system change if more models were introduced?
4. Can this method be applied to solve nonlinear systems?
5. What impact would it have if negative profits were expected for both models?

**Tip:** When solving simultaneous equations, it's often useful to simplify one equation before proceeding with elimination or substitution.

A computer dealer sells two models of computer A and B. In one particular week he sells 6 model A’s and 9 model B’s. The profit on these sales is P 585. He would make the same profit from selling 3 model of A’s and 2 model B’s. If the profit on a model A computer is P x and on a model B computer is P y,

a) Obtain two simultaneous equations in x and y to express these facts. [4 marks]

b) Solve these equations for x and y. [4 marks]

Learn how to solve a problem involving simultaneous equations to find the profit from selling two computer models, A and B. The problem includes setting up linear equations based on given conditions and using algebraic methods like substitution and elimination to solve for the variables. Suitable for students in Grades 9-11.

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