# Solving Systems in Grade 10

In grade 10 we worked on modelling word problems with equations, and solving two equations.

We started with the case of 8 apples and 3 oranges costing 12$, and 6 apples and 3 oranges costing 10.50$. We recognized that since in both situations the numbers of oranges is the same, that the price difference comes from the difference in apples.

2 apples account for a difference in price of 1.50$ which means that each apple costs 0.75$. We can then find the price of the oranges by using an equation.

6(0.75)+3(orange price)=10.50

3+3(orange price)=10.50

Subtract 3 from both sides

3(orange price)=7.50

Divide by 3

Orange price=2.50

Our next problem involved the mass of calculators and textbooks.

**5 calculators and 2 books have a mass of 5kg**

**10 calculators and 1 book have a mass of 6kg.**

The first step is to create a situation like with the apples and oranges (we want something the same between the 2 equations). We can double all quantities in the first equation to help.

Not all solution methods look the same, and that’s ok. We need to be sure we show what we are doing in a way others can follow it.

We tended to use algebra as our solving method. We can also solve quickly using desmos to graph the equations. The intersection point is the solution! Calculators have a mass of 0.467 kg and books have a mass of 1.33kg.

We finished with the same question that was tricky at the end of last class.