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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.

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• this impossible help me
• `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 :)`
• Trying to solve substitution problem.
x=3y-4
4y-2x=13
Do I need to use a negative fraction with X?
• 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.
• How do I solve an equation like this? X=y+8. And 3x-y=16 ? I am trying to solve for x and y
• x=y+8. Substitute that into the second equation. 3(y+8)-y=16. Distribute the 3.
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
• Can't we just put it into Desmos caculator and solve it that way?
• 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!
• how do I use substitution if the both equations have the y variable by itself?
• But how do I graph a system of equations when my x or y is a decimal?
• change it to a mixed number :DD
• How do i solve 5y(2y-3)+(2y-3)?
• 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
• -6x+6y=-24
-3x-12y=-12
• y=0
x=4
is one way to solve that problem
• how do I solve y=x^2-5x-14 and y=x-7
(1 vote)
• Since both equations give y explicitly in terms of x, we can easily substitute either equation into the other one, to get x^2-5x-14 = x-7.
Therefore, x^2-6x-7 = 0; (x-7)(x+1) = 0; x-7=0 or x+1=0; x=7 or x=-1.
Since y=x-7, it follows that y=0 when x=7, and y=-8 when x=-1.
So the solutions are the ordered pairs (7, 0) and (-1, -8).

Plugging in each of these two solutions into each of these two equations, we can verify that both solutions work (note that 7^2-5*7-14=0, 7-7=0, (-1)^2-5*(-1)-14 = -8, and -1-7 = -8).
It is also clear that there are no other solutions, because the graphs of a line and a parabola never intersect in more than two points.

Have a blessed, wonderful day!