Question

Suppose you got 8 mangoes and 3 apples for $18 and 3 mangoes and 5 apples for $14.50.
How would you write a formula to find out how much each apple and mango cost?

Answer 1

Define a variable and rewrite the question in mathematical manner.

x = mango price
y = apple price

8x + 3y = 18
3x + 5y = 14.5

Answer 2

Thank you but how do you define variables in this problem?

Answer 3

When you set up a variable in a problem-solving, first try to identify what you are interested in.

For example in this case, what are you interested in?

Yes, you are interested in the price of apple and mango.

But you don't know this yet. So you can name them whatever you want to call it. In this case I called them x and y.

Question told me that 8 mangos and 3 apples costed 18 bucks. This simply means price of 8 mangos and 3 apples yield 18.

which is 8x + 3y = 18.

Now in a similar manner, try to understand how the second equation got setup.

Answer 4

Ok, the second equation was 3x+5y=14.5 because there are three mangoes (x) and five apples (y) for a total of $14.5 but What I don't understand is how to solve an equation with two different variables

Answer 5

There's two ways of solving this, I'll explain the first method.

1)

Pick any one of the equation. For example, let's pick the first one.

8x + 3y = 18

You are interested in x and y. But you don't know them yet so let's start off by describing one of them. I'll chose x in this case (what you pick first doesn't matter).

Subtract 3y from both sides.

8x = 18 - 3y

And then divide 8 on both sides.

x = (18 - 3y)/8

What I did was algebraically manipulate the equation so that I can come up with an expression for x.

Now interestingly, the first and second equation have the same x. So what I say about x from the first equation must be true about the x in second equation.

Let's take a look at the second equation now. You said second equation is 3x+5y=14.5, which is exactly right.

3x+5y=14.5

Do you see what x or y is? Well, at least you know x because the first equation told us that x = (18 - 3y)/8! since x is the same as (18 - 3y)/8, plug it into the equation.

3((18 - 3y)/8) + 5y = 14.5

Now we can manipulate this equation to solve for y.

Multiply 8 on both sides

3(18-3y) + 40y = 116

Multiply out the 3(18-3y) expression.

54 - 9y + 40y = 116

Subtract 54 from both sides and add the y terms together.

31y = 62

Divide both sides by 31 so that we get an expression for y.

y = 2

We have found out that y is 2.

Now we can pick from first or second equation to find what x is (once again, it doesn't matter which one we pick).

Let's choose the first equation.

8x + 3y = 18

We know that y = 2, plug this into the equation.

8x + 6 = 18

Subtract 6 from both sides

8x = 12

Divide 8 on both sides

x = 12/8 = 1.5

Which concludes that x = 1.5, y = 2, and by how we defined the variables, it means that mango costs 1.5 bucks and apple costs 2 bucks.

This method is called "solving by substitution." We have 2 equations describing same 2 variables. So what we do is choose any equation and find an isolated expression for variable of our choice (like how I got x = (18 - 3y)/8).

We only get an algebraic expression from a single equation. But, since x is this equation should be the same as x in the other equation, we can replace the expression for x that we found from on equation to the remaining one.

Picking second equation and isolating y first to replace y in first equation also yields exactly the same result. So it will be a good practice for you to try solving this problem again with same method but starting from a different point.

I will now begin expressing the second method while you read this one.

Answer 6

2)

This method involves looking at the equation at the same time.

8x + 3y = 18
3x + 5y = 14.5

You should know that in an equation, if we add/subtract/multiply/divide same amount on both sides, then it should be fine.

For example,

8x + 3y + 10 = 18 + 10

We use this property to manipulate one of the equations to handle a remaining equation.


Let's focus our interest in y in this case.

First equation has 3 of y, and second equation has 5 of y.

We can try to manipulate both equations so that first and second equation has the same number of y.

That's pretty easy. Because multiplying 3 by 5 is equal to multiplying 5 by 3.

So what we do it multiply 5 on both sides of equation 1 and multiply 3 on both sides of equation 2.

This yields:

40x + 15y = 90
9x + 15y = 43.5

Now, notice how we can subtract the second equation from the first one!

let's subtract 9x + 15y from both sides of equation 1, this yields,

40x + 15y - (9x + 15y) = 90 - (9x + 15y)

Right?

But notice that 9x + 15y = 43.5, so we can replace the 9x + 15y on the right side with 43.5, which leads to

40x + 15y - (9x + 15y) = 90 - 43.5

Now let's add/subtract the terms. Notice how the y terms cancel each other out so that only x is remaining.

40x - 9x + 15y - 15y = 90 - 43.5

31x = 46.5

Divide both sides by 31,

x = 46.5/31 = 1.5

Now, we can just plug this into any of our initial equation just like what we did in last step of our first method, 'solving by substitution.'

8x + 3y = 18

8*(1.5) + 3y = 18

12 + 3y = 18

Subtract both sides by 6

3y = 6

Divide both sides by 3

y = 2

We solved the question and got the same answer using a different method.

This method is called "solve by elimination." What we did is manipulate 2 equations that we setup so that both of the equation have same number of x or y in it (I chose y this time but you can try this again with x, and still get the same answer).

Then subtracting (OR ADDING) so that one of the variables get "eliminated" to leave out only one term leads to solving one of the variables. And finding one leads to the other.

Any other questions?

Answer 7

Thank you so much, you really helped me!