Tutoring math, you notice practical problems that can be entertaining.  The math tutor presents one, highlighting the substitution method.

Example:  You have some oranges and a box.  When you weigh the box with three oranges in it, the mass is 1000g.  With ten oranges in it, the mass is 1770g.  Assuming the oranges are all the same, find the mass of an orange, as well as the mass of the box.

Solution:  We organize ourselves with a couple of “let” statements:

Let x=the mass of an orange.

Let y=the mass of the box.

Next, we translate The box with three oranges in it has mass 1000g into math, using our variables:

3x + y = 1000

Similar for With ten oranges in it, the box has mass 1770g:

10x + y = 1770

To solve the problem, an easy technique to use is substitution. To do so, we isolate y in the first equation (by subtracting 3x from both sides):

y = 1000 -3x

Now, we substitute 1000 – 3x for y in the second equation:

10x + 1000 – 3x = 1770

We take our new equation and simplify it:

7x + 1000 = 1770

Next, we subtract 1000 from both sides:

7x = 770

Dividing both sides by 7, we get

x = 110

Re-examining our let statements, we recall x is the mass of an orange. Now, to get y, we substitute 110 for x in either of our equations:

3(110) + y = 1000

330 + y =1000

Subtracting 330 from both sides, we get

y=670

So, the mass of an orange is 110g, while the box is 670g.

The substitution method is slick when isolating one of the variables leaves you with whole numbers. Otherwise, you might prefer the elimination method, which I will cover in a future post:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.