A Tutorial on Solving Equations Part 2

This is a continuation of Solving Equations Part 1.  As I have mentioned in that post, being able to solve equations is very important since it is used for solving more complicated problems (e.g. word problems).

In this post, we are going to solve a slightly more complicated equations. We already discussed 5 examples in the first post, so we start with our sixth example.

Example 6: 8x = 4x - 12

As I have mentioned in the previous examples, we need to isolate x on one side of the equation and all the numbers on the other side. Here, we decide to put all x‘s on the left hand side, so we remove 4x on the right hand side. To do this, we subtract 4x from both sides of the equation.

8x - 4x = (4x - 12) - 4x

Of course, 4x - 4x = 0, so, simplifying, we have

4x = -12

Then, we want to eliminate 4 on the left hand side. Since it is multiplication, we therefore divide both sides of the equation by 4.

\displaystyle \frac{4x}{4} = \frac{-12}{4}

Therefore, x = -3.

Example 7: x + 4 = 4x - 12

In this example, we want to avoid a negative x, so it is better to put all x‘s on the right hand side of the equation. This means that we have to eliminate x from the left hand side. So, we subtract x from the left hand side, and of course, the right hand side as well.

x + 4 - x = 4x -12 - x

4 = 3x - 12

Next, since we want to eliminate all the numbers on the right, the easiest to eliminate first is -12. To do this, we just add 12 on both sides of the equation.

4 + 12 = 3x - 12 + 12

16 = 3x.

Next, we only have one number on the right hand side which is 3. To eliminate it, we divide 3x by 3. Of course, we also need to divide the other side by 3.

\displaystyle \frac{16}{3} = \frac{3x}{3}

\displaystyle \frac{16}{3} = x

Therefore, the answer is x = \displaystyle \frac{16}{3}.

Notice also that we can add 2 and subtract x immediately resulting to x + 4 + 12 - x = 4x - 12 + 1 2 - x making the process faster. You will be able to discover such strategy on your own if you solve more equations.

Example 8: 2(2x - 3) = 5

In this example, we have the form a(b + c) in the left hand side of the equation. To simplify this, we simply distribute the multiplication of a over b + c. That is

a(b + c) = ab + ac.

This is called the distributive property of multiplication over addition.

So, solving the problem above, we have

2(2x) - 2(3) = 5

4x - 6 = 5

Adding 6 to both sides of the equation, we hhave

4x = 11

Dividing both sides of the equation by 4 we have

x = \frac{11}{4}.

Example 9: \frac{3x}{5} + 4 = 7

In equations with fractions, the basic strategy is to eliminate the denominator. In this example, the denominator is 5. Since \frac{3x}{5} means 3x divided by 5, we cancel out 5 by multiplying the equation by 5. Notice how 5 is distributed over the left hand side.

5 (\frac{3x}{5} + 4) = (5)(7) which is the same as

5 (\frac{3x}{5}) + 5(4) = 35.

Simplifying, we have 3x + 20 = 35.

Subtracting 20 from both sides, we have

3x = 15

Dividing both sides by 3, we have x = 5.

Example 10: \frac{3x + 3}{2} = x - 5.

We eliminate fraction by multiplying both sides of the equation by 2. That is

\displaystyle 2(\frac{3x + 3}{2}) = 2(x - 5)

In the left hand side, 2 cancels out 2, so only 3x + 3 is left. On the right hand side, we use distributive property.

3x + 3 = 2(x) - 2(5)

3x + 3 = 2x - 10

Subtracting 2x from both sides, we have

x + 3 = -10

Subtracting 3 from both sides, we have

x = -13

That’s it. In the next post, we solve more equations, particularly those that involve fractions.

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2 Responses

  1. March 13, 2014

    […] solve more complicated equations especially those that contain fractions.  The first part and the second part of this series discuss 10 sample equations. We start with the 11th […]

  2. July 24, 2016

    […] Solving Equations Part 2 […]

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