8.3 Solving linear equations using distributive property: a(x+b)=ca(x + b) = c

Back in previous chapters, we discussed applications of linear relations and operations of linear equations. We learned how to solve them using the four operations: addition, subtraction, multiplication, division and also by applying the distributive property. Now in this chapter, we will review all the aforementioned lessons, along with how to represent patterns in linear relations. If you’re given a linear equation and you’re asked to solve it using multiplication and division what should be you first step?

The first thing to do when solving for any variable like x, is to isolate it on one side of the equation. This means you have to be able to cancel out the numerical coefficient attached to it. To do that you need to use the numerical coefficient as the denominator in both sides of the equation. So if you’re given -9/7 = 3x, you would need to use 3 as the denominator of both 3x and -9/7 and simplify the equation so you would arrive at x = -3/7. 8.1 will be discussing more of these equations there are a lot more examples that we would be working on there.

In 8.2 we will be learning how to solve a two step linear equation. From the word two-step, this linear equation would require using two steps for us to solve for x. We have a sample equation 4/5 + 5/4x = 1/3, to solve for x we need to simplify one side of the equation to be able and then isolate x. In this case, we need to combine like terms 4/5 and 1/3 so we transpose 4/5 to the other side of the equation and use the opposite operation, so we get 5/4x = 1/3-4/5 or 5/4x = -7/15. By using cross multiplication we are able to isolate x and get the value -28/75.

In 8.3 we would be applying distributive property to solve for x. A brief review of what we have learned on linear equations in Grade 8, would tell us how to do this.

In the last part, we will be learning how to solve linear equations when variables are present in both sides by simply isolating the like terms with the variable in one side of the equation then combining them to get the final term before proceeding to the usual process of canceling out the numerical coefficient (if applicable).

Solving linear equations using distributive property: a(x+b)=ca(x + b) = c

When you see equations in the form of a(x+b) =c, you can remove the bracket and rewrite the equations into ax+ab =c using the distributive property. In this lesson, we will make use of this property to help us solve linear equations.

Lessons

  • 1.
    Solve.
  • 2.
    Find the value of xx.
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Solving linear equations using distributive property: a(x+b)=ca(x + b) = c

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