Example:\u00a0<\/strong> You have some oranges and a box.\u00a0 When you weigh the box with three oranges in it, the mass is 1000g.\u00a0 With ten oranges in it, the mass is 1770g.\u00a0 Assuming the oranges are all the same, find the mass of an orange, as well as the mass of the box.<\/p>\n Solution:<\/strong>\u00a0 We organize ourselves with a couple of “let” statements:<\/p>\n Let x=the mass of an orange.<\/p>\n Let y=the mass of the box.<\/p>\n Next, we translate The box with three oranges in it has mass 1000g<\/em> into math, using our variables:<\/p>\n 3x + y = 1000<\/p>\n Similar for With ten oranges in it, the box has mass 1770g<\/em>:<\/p>\n 10x + y = 1770<\/p>\n 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):<\/p>\n y = 1000 -3x<\/p>\n Now, we substitute 1000 – 3x for y in the second equation:<\/p>\n 10x + 1000 – 3x = 1770<\/p>\n We take our new equation and simplify it:<\/p>\n 7x + 1000 = 1770<\/p>\n Next, we subtract 1000 from both sides:<\/p>\n 7x = 770<\/p>\n Dividing both sides by 7, we get<\/p>\n x = 110<\/p>\n 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:<\/p>\n 3(110) + y = 1000<\/p>\n 330 + y =1000<\/p>\n Subtracting 330 from both sides, we get<\/p>\n y=670<\/p>\n So, the mass of an orange is 110g, while the box is 670g.<\/p>\n 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:)<\/p>\n