Main content

### Course: 6th grade (Eureka Math/EngageNY) > Unit 4

Lesson 6: Topic G: Solving equations- Variables, expressions, & equations
- Testing solutions to equations
- Intro to equations
- Testing solutions to equations
- Same thing to both sides of equations
- Representing a relationship with an equation
- One-step equations intuition
- Identify equations from visual models (tape diagrams)
- Identify equations from visual models (hanger diagrams)
- Solve equations from visual models
- Dividing both sides of an equation
- One-step addition & subtraction equations
- One-step addition equation
- One-step addition & subtraction equations
- One-step addition & subtraction equations
- One-step addition & subtraction equations: fractions & decimals
- One-step addition & subtraction equations: fractions & decimals
- One-step multiplication equations
- One-step division equations
- One-step multiplication & division equations
- One-step multiplication & division equations
- One-step multiplication & division equations: fractions & decimals
- One-step multiplication equations: fractional coefficients
- One-step multiplication & division equations: fractions & decimals
- Finding mistakes in one-step equations
- Finding mistakes in one-step equations
- Find the mistake in one-step equations

© 2024 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Testing solutions to equations

In this math lesson, we learn how to solve an equation with one variable. We test different values for x to see which one satisfies the equation. By substituting x with the given options, we find the correct solution that makes both sides of the equation equal. Practice makes perfect!

## Want to join the conversation?

- i think all this math is hurting my head(84 votes)
- ugh... same here i need a math buddy to help me(37 votes)

- Help! When I attempted to solve for x, I got 5x is equal to 4x. Here's how I did it.

5x-3=4x+3. -3 and +3 cancel out to 0, so I am left with5x=4x.(30 votes)- 5x-3=4x+3

5x=4x+6

x=6

Be careful with the signs of the constants or variables. The 3s definitely do NOT "cancel out." First I added 3 to both sides, then I subtract 4x from both sides. Hope this helps.(31 votes)

- So i did the equation, but im not sure i did it right.

5x-3=4x+3

+3 +3

5x = 4x+6

-4x -4x

x=6

Is this the right way to do it?(19 votes)- Yes. As shown in this video, you can also check your answer by substituting it back into the original equation. Simplify both sides and make sure they are equal.

5(6)-3 = 4(6)+3

30-3 = 24+3

27 = 27

They are equal, so you have a good solution.(17 votes)

- He makes everything look so easy(23 votes)
- I think your videos make my head spin and I have an A in math plus the highest A in the class. You need to just slow down and stop making people have to pause the video(15 votes)
- if ur a A in math then I think ur talking about regular 6th grade math(0 votes)

- 7+5(x-3)=22 if x=4,5,6 and 7 which is correct and could you give me an example(1 vote)
- I'll do the 4 as an example:

1) Swap out "x" and replace it with the given value: 7 + 5(4 - 3) = 22

2) Follow order of operations rules (PEMDAS) to simplify the left side

-- Do the work inside the parentheses: 7 + 5(1) = 22

-- Multiply: 7 + 5 = 22

-- Add: 12 = 22

3) Are the 2 sides equal? No. This means 4 is not a solution to the equation. If the 2 sides are equal, then the number would be a solution.

Hope this helps.(18 votes)

- If you don’t have choices then you can estimate.

Like 5x-3=4x+3 you can estimate to about 5.

Also, how do you simplify?(5 votes)- If you subtract 4x and add 3, you get 5x-4x-3+3=4x-4x+3+3, or x=6.

5(6)-3=27 and 4(6)+3=27, so it checks.(4 votes)

- 5(6)-3=4(6)+3

30-3=24+3

27=27

They are equal, so you have a good solution(5 votes) - Do you not work the problem so you get the answer rather than fill in possibilities? Ex:

5x-3=4x+3

5x-3+3=4x+3+3

5x=4x+6

5x-4x=4x-4x+6

×=6(3 votes)- This video was about testing solutions.

-- You could be given a problem that asks you: is x=7 a solution?

-- You could have found a possible solution like you did in your work above and you want to verify that you have the correct solution.

In both of these situations, you would use the techniques shown in the video.(5 votes)

- Did you watch the video? Did you read the other questions and responses. Multiple responses provide step-by-step instructions. Here is one that I posted 16 days ago:

If someone gives you an algebraic equation and asks if a specific value is a solution or not, here's what you do.

1) Replace each instance of the variable in the equation with the value you were given.

2) Follow order of operations rules (PEMDAS) for each side of the equation. Do the math until you have one number on both sides.

3) If the 2 sides are equal (they have the same number), then the value you were given is a solution to the equation. If the 2 sides are not equal, then you either have a math error or the given value is not a solution to the equation.

Hope this helps.

P.S. In future, look at the other info available first. Then try to ask a more specific question about what you don't understand. It will help you learn faster.(9 votes)

## Video transcript

- [Voiceover] So we've
got an equation here, it says five times x minus three is equal to four times x plus three. So, what we want to do
is we want to figure out an x that satisfies this,
so there's some number that if I take five,
multiply it by that number, subtract three from it, that's going to be the same thing as if I
take four times that number and add three to it. And, before we go into how to
solve these types of things, let's just first see
if we can test whether something does satisfy this equation. And so, I have three options
here, I have x could be equal to five, x could be equal to six, and x could be equal to seven. And your goal is to pause
this video and figure out which of these x's
satisfies this equation, which of these values would
make this equation be true. So, I'm assuming that you have tried that, so let's work through each
of them, step one by one. So let's see this first
one: if x is equal to five, then in order for this to
be true, five time five, right, five times x, so
five times five minus three needs to be equal to four
times, everywhere we see an x we're going to put a five there, four times, actually
let me do it this way. Let me just color code it. So this is the same thing as saying five times five minus three, let me do
that in that same color, minus three, needs to be
equal to four times five, four times five plus three, plus three. Color changing is hard. Plus three. Now, is this true? Let's see, five times five
is 25, it's going to be 25 minus three, needs to
be equal to 20 plus three. 25 minus three is 22,
needs to be equal to 23. No, this is not true. So, x does not equal five,
so this is not a solution. Let's try x equals six. So, once again, we're going
to do five times our x, which is going to be six,
actually let me just write it out, minus three needs to be
equal to four times our x, plus three, and in this case our x is six, so it's going to be five
times six minus three needs to be equal to four
times six plus three. What's five times six? Well, it's 30 minus
three, needs to be equal to four times six is
24, and then plus three. Well, this is true, 30 minus three is 27, which is indeed equal to 24
plus 3, it's equal to 27. So x equals 6 does satisfy our equation, it is a solution, and actually
as we'll see in the future, the solution to this
equation right over here. X equals six satisfies this. Now, just for good
measure, let's just varify that x equals seven will not satisfy. So I'm going to move this up a little bit. So if x is equal to
seven, we're going to get five times seven minus
three needs to be equal to four time seven plus three. And so, we're going to
get, and in all these cases we do the multiplication
first, order of operations, and it's very clear
when you see it kind of in the algebraic notation up
here, so we're going to do 35 minus three needs to
be equal to 28 plus three, 35 minus three is 32, 28 plus three is 31, these do not equal each other. So this is not a solution
to our original equation.