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### Course: 8th grade > Unit 4

Lesson 1: Intro to systems of equations# Testing a solution to a system of equations

Sal checks whether (-1,7) is a solution of the system: x+2y=13 and 3x-y=-11. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- What is system?(33 votes)
- A "system" of equations is a set or collection of equations that you deal with all together at once.(11 votes)

- would this work for a quadratic equation?(33 votes)
- Yes because you will want you to check to see if you have the right solution.(9 votes)

- Does a single linear equation with two or more unknowns always have infinitely many solutions(14 votes)
- Yes, that's right. You could choose whatever values you like for all but one of the variables, and then final variable can always be made to fit.

For example, if you had the equation`ax + by + cz = k`

, then whatever you pick for`y`

and`z`

, you can solve for`x`

to get`x = (k - by - cz)/a`

, and the equation will be satisfied.(24 votes)

- im stupid i dont get this(9 votes)
- Bro chill. A solution of an equation is when both sides (i.e., LHS and RHS) become equal. What do you need to do to make both sides equal? Well, you need to find some values for X and Y so that they become equal when you plug X values wherever X and Y are.

Here, some of the solutions are given, but we need to check after plugging them in it makes both sides of the equation equal. (like 1 = 1 , 2 = 2, BUT if you get 1 = 2, or 3 = 4 it is clear that it is false and hence the values of X or Y or both are wrong and hence, not the solution[s] )(14 votes)

- Would this work for quadratic equations?(7 votes)
- Since you are testing the point for each equation independent of each other, it would work for any function. If you have two quadratic equations, there is also a possibility of having two different intersections, not just one.(14 votes)

- I have perfectly parallel lines, so is there a solution? I can't figure out this problem.(7 votes)
- Parallel lines will never cross so a system of parallel lines will have no solution.

As a reminder, parallel lines have the exact same slope.(14 votes)

- is it just me or am i just really dumb? nothing makes sense(8 votes)
- By now you should be familiar with the concept of testing solutions to equations by using substitution. If you are asked if a point is a solution to an equation, we replace the variables with the given values and see if the 2 sides of the equation are equal (so is a solution), or not equal (so not a solution).

A solution to a system of equations means the point must work in both equations in the system. So, we test the point in**both**equations. It must be a solution for both to be a solution to the system.

Hope this helps.(4 votes)

- can u make an example more easier(5 votes)
- The example in the video is about as simple as it gets. Neither equation has fractions or decimals.

The video is show you how to determine if an ordered pair (a point) is a solution to a system of equation. Sal has one point that he is testing to see if it is a solution to the system. In order for this to be true, the point must work in both equations (i.e., the 2 sides of each equation come out equal). He does the test by substituting the values from the ordered pair into each equation and simplifying.

-- The point works in the 1st equation.

-- The point did not work in the 2nd equation.

This tells us the point in on the line created by the first equation, but it is not a point on the line created by the 2nd equation. Remember, to be solution to the system, the point must work for both equations. Since it didn't, the point is not a solution to the system.

Hope this helps.(7 votes)

- i dont understand math im confused(7 votes)
- Math uses knowlegde to percilcoly explain how to solve problems(0 votes)

- Can there be systems of equations with three or more equations?(4 votes)
- Absolutely there can be. If there were three or more equations in the system, you can solve for even more unknowns.

Always remember to solve every variable's value in terms of`1`

of the variables.

For simplicity, I've made up a`3`

-equation system that already has`2`

variables solved in terms of`1`

variable in particular -`y`

.`{`

(1) x = y + 1

(2) z = y/2

(3) x + y + z = 11

}

Plugging in (1) and (2) into (3):`(y+1) + (y) + (y/2) = 11`

`y(5/2) + 1 = 11`

`y(5/2) = 10`

`y = 4`

Then, using (1) and (2) to find the remaining variables:`x = (4) + 1 = 5`

`z = (4)/2 = 2`

So`x = 5`

,`y = 4`

and`z = 2`

. This satisfies all equations in the system!

Happy learning.(5 votes)

## Video transcript

Is negative 1 comma 7 a
solution for the system of linear equations below? And they give us
the first equation is x plus 2y is equal to 13. Second equation is 3x minus
y is equal to negative 11. In order for negative
1 comma 7 to be a solution for the
system, it needs to satisfy both equations. Or another way of thinking about
it, x equals 7, and y-- sorry, x is equal to negative 1. This is the x coordinate. X equals negative 1, and
y is equal to 7, need to satisfy both of
these equations in order for it to be a Solution. So let's try it out. Let's try it out with
the first equation. So we have x plus
2y is equal to 13. So if we're thinking
about that, we're testing to see if when x
is equal to negative 1, and y is equal to 7,
will x plus 2y equals 13? So we have negative
1 plus 2 times 7-- y should be 7-- this
needs to be equal to 13. And I'll put a
question mark there because we don't
know whether it does. So this is the same
thing as negative 1 plus 2 times 7 plus 14. That does, indeed, equal 13. Negative 1 plus 14, this is 13. So 13 does definitely equal 13. So this point it does, at least,
satisfy this first equation. This point does sit on the
graph of this first equation, or on the line of
this first equation. Now let's look at
the second equation. I'll do that one in blue. We have 3 times negative
1 minus y, so minus 7, needs to be equal
to negative 11. I'll put a question
mark here because we don't know whether
it's true or not. So let's see, we have 3 times
negative 1 is negative 3. And then we have minus 7 needs
to be equal to negative 11-- I put the question mark there. Negative 3 minus 7,
that's negative 10. So we get negative 10
equaling negative 11. No, negative 10 does
not equal a negative 11. So x equaling negative
1, and y equaling 7 does not satisfy
the second equation. So it does not sit on its graph. So this over here is not
a solution for the system. So the answer is no. It satisfies the
first equation, but it doesn't satisfy the second. In order to be a
solution for the system, it has to satisfy
both equations.