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- EQAO Grade 9 Foundations of Math
- Solving Linear Equations

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Get Started Now- Intro Lesson17:59
- Lesson: 1a3:30
- Lesson: 1b3:04
- Lesson: 1c4:00
- Lesson: 1d2:42
- Lesson: 2a2:28
- Lesson: 2b6:38
- Lesson: 3a1:38
- Lesson: 3b3:09
- Lesson: 3c1:29
- Lesson: 3d3:01
- Lesson: 42:38

Solving two-step linear equations literally means solving the equations in two major steps. First, we need to isolate the unknown "x" to one side of the equation. You can then solve the "x". To complete these two steps, you may need to perform addition, subtraction, division, multiplication, cross multiplication, and so on. When the equation has fractions, you may also need to find the common denominator before proceeding further. Seems complicated? No worries! You will learn all the tricks in this section.

Basic Concepts:Model and solve one-step linear equations: $ax = b$, $\frac{x}{a} = b$, Solving two-step linear equations using addition and subtraction: $ax + b = c$, Solving two-step linear equations using multiplication and division: $\frac{x}{a} + b = c$,

Basic Concepts:Solving multi-step linear inequalities, Introduction to linear equations, Applications of polynomials,

- Introductiona)How to turn a word problem into an equation?

• ex. 1: "revenue" problem

• ex. 2: "area" problem - 1.Solve.a)$\frac{4}{5} + \frac{5}{4}x = \frac{1}{3}$b)$\frac{3}{4} + 2x = 5\frac{1}{3}$c)$\frac{2}{3} - \frac{x}{2} = \frac{4}{7}$d)$- 3\frac{3}{4} = - 6\frac{1}{4} + \frac{1}{8}x$
- 2.Solve.a)$- 0.05 - \frac{x}{{2.6}} = - 0.03$b)$\frac{x}{{ - 2.14}} + 0.86 = 6.32$
- 3.Solve.a)$3.07 = 0.3x - 4.6$b)$\frac{7}{9} = \frac{7}{8} - \frac{x}{9}$c)$- 1.8 = 4.5 + \frac{x}{{2.3}}$d)$3\frac{1}{3} + 2\frac{1}{9}v = - \frac{4}{9}$
- 4.The number of hours Peter exercised in May is 3.5 hours less than one fourth of the number of hours John exercised in the same month. Peter had 15.8 hours of exercise in May. How many hours of exercise did John have in May?

We have over 620 practice questions in EQAO Grade 9 Foundations of Math for you to master.

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