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# Systems of equations with substitution: 2y=x+7 & x=y-4

When solving a system of equations using substitution, you can isolate one variable and substitute it with an expression from another equation. This will allow you to solve for one variable, which you can then use to solve for the other. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

• this impossible help me
(8 votes)
• `Ok, let's take the same problem and break it down, very carefully. 2y = x + 7 &x = y - 4so the second equation above denotes that x_ correlates to _y - 4. So, let us substitute y - 4 on the top of the equation replacing the position of the value x. Making it 2y = (y-4) + 7. Let us find the value of y. 2y = y + 3 <-- Here, we have combined like terms first. (negative 4 plus 7). *-y + 2y = y - y + 3* <-- Now, we have subtracted y from one side to the other (variable) side of the equation. y = 3 <-- Now, we found the value of the variable, y_. Which is _3. Let us now substitute the value of y, for the second term first -x = y - 4x = 3_ - 4 <-- Replaced the value of _y with the constant value 3_. x = - 1 <-- We found the value of *_x*!!Now, x is equal to negative 1 and y is equal to 3. To prove that these answers as valid, try to substitute into both equations. By substituting these x_ and _y values, we should be able to view both sides of the equation, as true. Thanks :)`
(32 votes)
• Trying to solve substitution problem.
x=3y-4
4y-2x=13
Do I need to use a negative fraction with X?
(16 votes)
• Just substitute the value of x in the first equation into the other equation and solve for y.
4y - 2x = 13
4y - 2(3y - 4) = 13
Once you have solved that, substitute the value of y back into the first equation and solve for x.
(14 votes)
• How do I solve an equation like this? X=y+8. And 3x-y=16 ? I am trying to solve for x and y
(10 votes)
• x=y+8. Substitute that into the second equation. 3(y+8)-y=16. Distribute the 3.
3y+24-y=16. add like terms.
2y+24=16. subtract 24 from both sides.
2y=-8. divide by 2
y=-4. Now that you know y=-4, plug that in the first equation.
x=-4+8.
x=4
(12 votes)
• Can't we just put it into Desmos caculator and solve it that way?
(8 votes)
• Why does substitution work?
The two equations are separate equations. Why does this work? The graph seems easy to get, but this is a bit harder.
I am having a hard time visualizing substitution or just understanding why it works. thanks!
(5 votes)
• how do I use substitution if the both equations have the y variable by itself?
(4 votes)
• you don't! you already have your answer there.
(2 votes)
• But how do I graph a system of equations when my x or y is a decimal?
(2 votes)
• change it to a mixed number :DD
(3 votes)
• at that was fascinating
(3 votes)
• How do i solve 5y(2y-3)+(2y-3)?
(2 votes)
• 5y(2y-3) + (2y-3) = (2y-3) (5y+1) = 10y^2 -13y - 3

Here, I took the (2y-3) as common factor using the additive distributive identity we had studied earlier. Hope you understood.

Another method will be 5y(2y-3)+(2y-3) = 10y^2-15y+2y-3 = 10y^2-13y-3
(2 votes)
• -6x+6y=-24
-3x-12y=-12
(2 votes)
• y=0
x=4
is one way to solve that problem
(2 votes)

## Video transcript

Use substitution to solve for x and y. And we have a system of equations here. The first equation is 2y is equal to x plus 7. And the second equation here is x is equal to y minus 4. So what we want to do, when they say substitution, what we want to do is substitute one of the variables with an expression so that we have an equation and only one variable. And then we can solve for it. Let me show you what I'm talking about. So let me rewrite this first equation. 2y is equal to x plus 7. And we have the second equation over here, that x is equal to y minus 4. So if we're looking for an x and a y that satisfies both constraints, well we could say, well look, at the x and y have to satisfy both constraints, both of these constraints have to be true. So x must be equal to y minus 4. So anywhere in this top equation where we see an x, anywhere we see an x, we say well look, that x by the second constraint has to be equal to y minus 4. So everywhere we see an x, we can substitute it with a y minus 4. So let's do that. So if we substitute y minus 4 for x in this top equation, the top equation becomes 2y is equal to instead of an x, the second constraint tells us that x needs to be equal to y minus 4. So instead of an x, we'll write a y minus 4, and then we have a plus 7. All I did here is I substituted y minus 4 for x. The second constraint tells us that we need to do it. y minus 4 needs to be equal to x or x needs to be equal to y minus 4. The value here is now we have an equation, one equation with one variable. We can just solve for y. So we get 2y is equal to y, and then we have minus 4 plus 7. So y plus 3. We can subtract y from both sides of this equation. The left hand side, 2y minus y is just y. y is equal to-- these cancel out. y is equal to 3. And then we could go back and substitute into either of these equations to solve for x. This is easier right over here, so let's substitute right over here. x needs to be equal to y minus 4. So we could say that x is equal to 3 minus 4 which is equal to negative 1. So the solution to this system is x is equal to negative 1 and y is equal to 3. And you can verify that it works in this top equation right over here. 2 times 3 is 6 which is indeed equal to negative 1 plus 7. Now I want to show you that over here we substituted-- we had an expression that, or we had an equation, that explicitly solved for x. So we were able to substitute the x's. What I want to show you is we could have done it the other way around. We could have solved for y and then substituted for the y's. So let's do that. And we could have substituted from one constraint into the other constraint or vice versa. Either way, we would have gotten the same exact answer. So instead of saying x is equal to y minus 4, in that second equation, if we add 4 to both sides of this equation, we get x plus 4 is equal to y. This and this is the exact same constraint. I just added 4 to both sides of this to get this constraint over here. And now since we've solved this equation explicitly for y, we can use the first constraint, the first equation. And everywhere where we see a y, we can substitute it with x plus 4. So it's 2 times-- instead of 2 times y, we can write 2 times x plus 4. 2 times x plus 4 is equal to x plus 7. We can distribute this 2. So we get 2x plus 8 is equal to x plus 7. We can subtract x from both sides of this equation. And then we can subtract 8 from both sides of this equation, subtract 8. The left hand side, that cancels out. We're just left with an x. On the right hand side, that cancels out, and we are left with a negative 1. And then we can substitute back over here we have y is equal to x plus 4, or so y is equal to negative 1 plus 4 which is equal to 3. So once again, we got the same answer even though this time we substituted for y instead of substituting for x. Hopefully you found that interesting.