When it comes to more complicated equations, we want to use this outline of the steps to follow in the solution process.
Step 1: Simplify each side of the equation as much as possible.
That usually involves using the distributive property, commutative property, associative property.
Step 2: Collect all the variable terms on one side of the equation and all the constant terms on the other.
As you'll see, that step involves the addition property.
Step 3: Use the multiplication property to get this just one x alone on one side of the equation.
Step 4: Check your solution in the original equation.
So, we simplify each side; use the addition property to separate the variable terms to one side, constant terms to the other;
then, we use the multiplication property. I'll show you what I mean. Here, we have an equation: 6(x - 3) = -6.
We need to start by separating the terms over here by using the distributive property. So, 6(x) = 6x; 6(3) = 18.
So, that's what I do first. Now, I'm going to use the addition property, this is a variable term, that's a constant term, and that's a constant term.
We want all the variable terms over on the left side and all the constant terms on the right side.
So, I need to add the opposite of this term to both sides: 6x - 18 + 18 = -6 + 18.
So, you can see there, I've added 18 to both sides. 6x + (-18) + 18 is simply 6x; and the -6 + 18 is going to be 12.
Next, I use the multiplication property. (1/6)(6x) on the left side; (1/6)(12) on the right side. (1/6)(6x) = x. (1/6)(12) = 2.
So, I get x = 2 for that solution. Now, before I leave this problem, I want to check this solution x = 2 back in this equation.
So, I'll substitute x = 2 into this equation. I get 6(2) - 3 = -6; 2 - 3 = -1.
And I end up with -6 = -6. So, you can see I get a true statement; that means that I was correct in my solution process
and got x = 2. If I had found out that I got a false statement right here, I would go back and check my work to see that I,
to see where I had made a mistake and gotten the wrong solution. Simplify each side first; addition property; multiplication property
there's your solution. Check it. Here's our next problem: 7y - 3 = 4y - 15;
variable term, variable term, constant term, constant term. Variable terms have the variable in them; constant terms do not.
So, what I want to do is get all the variable terms on one side and all the constant terms on the other.
So, I'm going to start by getting rid of this 4y that's on the right side of the equation. To do that,
I'm going to add its opposite, -4y, to both sides. So, you can see that here I'm adding a -4y to both sides of the equation.
It's okay to do that with the addition property of equality; I can add the same quantity to both sides of the equation
any time it's convenient for me. Now, 7y + (-4y) = 3y - 3 = 4y + (-4y) = 0.
So, see I've succeeded in getting rid of the y's that were on the right side of the equation. All that's left is negative 15.
Now, I'll add 3 to both sides, again, using the addition property.
And so, I have 3y + 0, which is 3y, is equal to -15 + 3, which is -12. Now, 3y is equal to -12.
I multiply both sides by 1/3. And maybe you can see that solution already; 3 times what is negative 12?
Of course, the answer has to be y is equal to -4. So, y = -4 is the solution to that equation.
It took me two applications of the addition property to get all the variable terms on one side and all the constant terms grouped together,
on the other side. Once that was done, it's a simple application of the multiplication property to get that solution.
Here's another example: 2(3x) - 6 + 1 = 7. Let's multiply using the distributive property.
2(3x) = 6x, - 2(6) is 12, + 1 = 7. 6x + (-12) + 1 is 6x - 11 = 7.
So, there it took me three lines to simplify both sides as much as possible.
I don't want to use that addition property or multiplication property until I have simplified each side as much as I can.
Now, I'll isolate x, all the variable terms with x in them, on the left side and put the constant terms on the other side.
6x - 11; that's right, I have to add 11 to both sides. 6x + 0 = 6x. Now, 6(x) = 18.
If I multiply both sides by 1/6, I'll get x = 3. 6(3) = 18. And I can either solve that by inspection, just by looking at it;
or I could multiply both sides by 1/6 and get that x = 3. This will check if I put it back into the original equation. Let's go on to our next problem,
a little more complicated looking. We're going to simplify each side as much as possible. So, let's start with
let's see, there's not much to do on the left side, so I'll just rewrite it: 9x - 6.
Here, I'm going to use the distributive property; and remember, I want to take the negative sign with the 3 when I multiply.
So, -3x; and then, -3(2) = -6, -24.
9x - 6 = -3x minus, that's a -6 plus a -24, which is -30. Let's get all our variable terms on the left side.
So, I'll start by getting rid of the variable term that's on the right side. I'll do that by adding its opposite to both sides:
9x + 3x + 12x, - 6, = 0 + (-30), which is just -30. Now, I'll add 6 to both sides of the equation:
12x - 6 + 6 = -30 + 6. 12x + 0 = 12x; -30 + 6 = -24.
Multiply both sides by 1/12, or you can probably solve this by inspection. There it is,
showing the multiplication property, 1/12 of 12x is simply x; and 1/12 of -24 = -2. So, -2 is my solution for that equation.
So, it took three steps right here to simplify both sides as much as possible;
then, two applications of the addition property to get all the variable terms on one side and constant terms on the other side.
Then, it was simply a matter of using the multiplication property to get just 1x. The next two equations we solve involve fractions;
and we're going to alter our steps in our solution process just a little bit. Here, I have x/5 - x = 4. Now,
I could solve this problem right here by working with the fractions. But sometimes, it's easier if we get rid of the fractions in the beginning.
So, I look at all three terms here as if they're fractions. This has a denominator of 5, denominator of 1, denominator of 1.
The least common denominator for these fractions is the number 5. I'm going to multiply the left side by 5,
and I'm going to multiply the right side by 5. So, I use my multiplication property a little bit out of order here:
I multiply both sides by the least common denominator first. But the reason why I do it is to clear the equation of fractions.
So, 5(x/5) is just x; 5(x) = 5x; and then, 5(4) = 20.
So, you can see, 1x - 5x = -4x, equal 20. And the -4 times what is 20? x turns out to be -5. So,
this method of taking, finding the least common denominator first and multiplying both sides by it gives us an equation that's free of fractions.
And that usually is a little easier to solve. I didn't show all the steps here in the solution process.
But you can see that this is not a difficult equation to solve. Let's try one more of those least common denominator problems:
1/x - 1/2 = -1/4. Least common denominator is equal to 4x; least common denominator is 4x.
So, I'm going to multiply the left side by 4x; and then, I'm going to multiply the right side by 4x. 4x(1/x) would be equal to 4,
the x's divide out. 4x(1/2) will be 2x, the 2's divide out. And then, -1/4(4x) will be equal to -x.
So, I get 4 - 2x = -x when I multiply both sides of this equation by the least common denominator 4x.
That gives me a simpler equation to work with, right here. Let's get all the variable terms on the right side, this time.
I'll do that by adding 2x to both sides. When I do that, the left side is 4 and the right side is simply -x + 2x, which is x.
So, my solution just falls out real quickly when I use that addition property right there. So, these are the problems
that you're going to encounter last in the problem set. They involve a least common denominator, and
it requires that you use the multiplication property a little bit out of order from the way that you're using it in the problems that come before these.