# Beth and Alex have 155p between them.Beth has 15p more than Alex.Beth have x p and Alex have y p. How do I formulate a simulteneous equation

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I'm not sure this is a 'simultaneous equation' but this is how i'd solve it.

'p' is just the unit, not a variable, so ignore 'p' for now.

Beth and Alex have 155 between them.

Beth has 15 more than Alex. Beth's amount is represented by x and Alex's amount is represented by y. Since Beth has 15 more than Alex, instead of using x for Beth, use (y+15). Now, Alex is y and Beth is (y+15).

The first equation would be: x + y = 155.

Substituting (y+15) for Beth's x, the new equation is:

y + (y+15) = 155

Solve for y: 2y + 15 = 155

2y = 140

y = 70

Beth is x, or (y +15), so x = (70+15)

x = 85

Then you can add 'p'. Alex has 70p and Beth has 85p giving 155p in total.

Simultaneous equations are equations that have these following characterisitics:

- They have multiple variables
- The variables must be solved for simultaneously.

In the situation you have provided your variables are x (Beth) and y (Alex). Think about what you know and what equations you can make:

Between them they have 155p. You can express this as x + y = 155 because Beth's money plus Alex's is 155p.

But you also know how to compare the amounts they have. Beth has 15p more so x = y + 15 because Beth's amount of money (x) is Alex's (y) plus 15.

So your equations are

x + y = 155

and

x = y + 15

So, what you're trying to do is ask yourself "what do I know about the relationships between the variables?" Once you figure that out, you ask "how do I express that relationship mathematically?"

I hope that helps you understand the process of getting the right answer.

From the data, the p's with Beth and Alex are xp and yp and their total is 155p. This is formulated as:

xp+yp = 155p .............(1)

Beth's xp is 15p more than Alex's yp. So,

xp-yp = 15P.................(2)

The two equations at (1) and (2) are the simultaneous equations which could now be solved for x and y and their values could be expressed in terms of p's.