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# Systems of equations with substitution: y=-5x+8 & 10x+2y=-2

Learn to solve the system of equations y = -5x + 8 and 10x + 2y = -2 using substitution. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

• What should u do if they ask u to verify it without a graph ?
• Put both equations in slope y-intercept form
If both have the same slope and different
y-intercept then these lines are parallel
and the system has no solution
• To extend this concept, can you solve a linear system of equations with 3 unknowns?
``x + y + z = 4x - y + z = 6-x + y + z = 0``

The three equations are planes in Euclidean three-space (https://en.wikipedia.org/wiki/Three-dimensional_space).
If a solution exists, it's the point (x,y,z) at which all three planes intersect.
• x + y + z = 4 (1)
x - y + z = 6 (2)
-x + y + z = 0 (3)
To find the point in which all three planes intersect, we first find a common coordinate within two equations by elimination or substitution. From there, we use a different pair of equations to find a different coordinate, then plug those two coordinates into all three equations to find and confirm the final coordinate (equations are numbered for clarification):
x - y + z = 6 (2)
-x + y + z = 0 (3)
2z = 6 (elimination)
z = 3

x + y + z = 4 (1)
x - y + z = 6 (2)
2x + 2z = 10 (elimination)
x + z = 5
x + 3 = 5 (We know that z = 3, so we simply plug that in.)
x = 2

x + y + z = 4 (1)
x - y + z = 6 (2)
-x + y + z = 0 (3)

2 + y + 3 = 4
2 - y + 3 = 6
-2 + y + 3 = 0

5 + y = 4
-y + 5 = 6
y + 1 = 0

y = -1
-y = 1
y = -1

y = -1
y = -1
y = -1

That means the solution to this system is (2 , -1 , 3).
• How would I be able to solve the variable for x and y I'm the following equation
x+3y=12
x-y=8
• To solve a system like this, you have two possible methods you can use: elimination or substitution. Substitution will be easy here since you don't have coefficients on several of the variables.
1) pick an equation and isolate a variable
x - y = 8 ---add y to both sides
x = y + 8
2) put this expression in place of the x in the other equation
x + 3y = 12
(y + 8) + 3y = 12 -- group like terms
(y + 3y) + 8 = 12 -- add
4y + 8 = 12 -- subtract 8 from both sides
4y = 4 -- divided by 4
y = 1
3) put this back into the equation from step 1 to find x
x = y + 8
x = 1 + 8
x = 9

• Solve the set of linear equations :
3x-4y=1
4x -3y=6
• How would you solve this problem?
4x + 5y = 11
y = 3x - 13
• Once you find x, you will need to substitute back into one of the equations to find y as well.
y = 3(4) - 13 = -1
Solution is x = 4, y = -1 or (4,-1)
• How would I be able to solve the variable x and y on the following equation

4x-3y=12
x+2y=14
• Luzlilvillalobos,
4x-3y=12
x+2y=14
To solve by substitution, you first need to isolate one of the variables in one equation by itself on one side the equation
Let's isolate the x in the second equation.
x+2y=14 Subtract 2y from both sides
x=14-2y
Now you can substitute (14-2y) for the x in the other equation.
4x-3y=12 Put (14-2y) in for the x
4(14-2y) - 3y = 12
Now you can solve for y
And then put that answer in for y in either original equation and solve for x.

I hope that helps make it click for you
• How would solves equations with squares? For example:

xy=12
x^2+y^2=40
• So if two equations have the same slope then they are parallel and will never intersect? Could you therefore refer to the value on the X coefficients as proof without substitution that a system of equations in slope intersect form has no solution? Like a shortcut.
• Yes, if you can write each equation in slope-intercept form and show that the coefficients on x are the same while the constant terms are different, this would prove that the lines never intersect (no solution).

Have a blessed, wonderful day!
• How do I solve y=2x+7 and y=9x+8 using substitution?
• solve for x and y by substitution method
x+y= -1
x^2+y^2=13
• First, solve the first equation for one of the variables, say... x
x = -y - 1
Now, substitute the right-hand side of the above equation in to every x in the second equation
(-y - 1)² + y² = 13
(-y - 1)(-y - 1) + y² = 13
y² + 2y + 1 + y² = 13
2y² + 2y -12 = 0
y² + y - 6 = 0
(y + 3)(y - 2) = 0
y + 3 = 0
y = -3
x = -(-3) - 1
x = 3 - 1
x = 2
y - 2 = 0
y = 2
x = -2 - 1
x = -3
So, the points (-3, 2), and (2,-3) are solutions
(1 vote)

## Video transcript

Use substitution to solve for x and y. And they give us a system of equations here. y is equal to negative 5x plus 8 and 10x plus 2y is equal to negative 2. So they've set it up for us pretty well. They already have y explicitly solved for up here. So they tell us, this first constraint tells us that y must be equal to negative 5x plus 8. So when we go to the second constraint here, every time we see a y, we say, well, the first constraint tells us that y must be equal to negative 5x plus 8. So everywhere we see a y, we can substitute it with negative 5x plus 8. Because that's what the first constraint tells us. y is equal to that. I don't want to be repetitive, but I really want you to internalize that's all it's saying. y is that. So every time we see a y in the second constraint, we can substitute it with that. So let's do it. So the second equation over here is 10x plus 2. And instead of writing a y there, and I've said it multiple times already, we can write a negative 5x plus 8. The first constraint tells us that's what y is. So negative 5x plus 8 is equal to negative 2. Now, we have one equation with one unknown. We can just solve for x. We have 10x plus. So we can multiply it. We can distribute this 2 onto both of these terms. So we have 2 times negative 5x is negative 10x. And then 2 times 8 is 16. So plus 16 is equal to negative 2. Now we have 10x minus 10x. Those guys cancel out. 10x minus 10x is equal to 0. So these guys cancel out. And we're just left with 16 equals negative 2, which is crazy. We know that 16 does not equal negative 2. This is an inconsistent result. And that's because these two lines actually don't intersect. And we could see that by actually graphing these lines. Whenever you get something like some number equalling some other number that they're clearly not equal to, that means it's an inconsistent result, It's an inconsistent system, and that these lines actually don't intersect. So let me just graph these just to make it clear. This first equation is already in slope y-intercept form. So it looks something like this. That's our x-axis. This is our y-axis. And it's negative 5x plus 8, so 1, 2, 3, 4, 5, 6, 7, 8. And then it has a very steep downward slope. Every time you move forward 1, you have to go down 5. So it looks something like that. That's this first equation right over there. The second equation, let me rewrite it in slope y-intercept form. So it's 10x plus 2y is equal to negative 2. Let's subtract 10x from both sides. You get 2y is equal to negative 10x minus 2. Let's divide both sides by 2. You get y is equal to negative 5x, negative 5x minus 1. So it's y-intercept is negative 1. It's right over there. And it has the same slope as this first line. So it looks like this. It's parallel. It's just shifted down a bit. So it just looks like that. So they're parallel lines. They have the same slope, different y-intercepts. We get an inconsistent result. They don't intersect. And the telltale sign of that, when you're doing it algebraically, is you get something wacky like this. This is why it's called inconsistent. It's not consistent for 16 to be equal to negative 2. These don't intersect. There's no solution to both of these constraints, no x and y that satisfies both of them.