- Systems of equations with substitution: 2y=x+7 & x=y-4
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Systems of equations with substitution: 2y=x+7 & x=y-4
Learn to use substitution to solve the system of equations 2y = x + 7 and x = y - 4. Created by Sal Khan and Monterey Institute for Technology and Education.
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- Trying to solve substitution problem.
Do I need to use a negative fraction with X?(10 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.(11 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(7 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.
How do u solve this problem by using sub my teacher did this in class using sub but thier is no y= or x= so i have to get coordinates like : (-9,2) or something like that how would you do that i know how to solve basic sub and what would this method be called?
Thank You(2 votes)
- You must first solve one of the equations for one of the variables. In this example, unfortunately, there is not a variable that you can isolate without getting fractions. For example, you might solve the first equation for x to get:
x = (-3/2)y + 5
You must substitute that into the other equation:
3x + 4((-3/2)x+5) = 8
Simplify and solve for x! Luckily, the fraction will disappear when you distribute the 4 into the parentheses.(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(3 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!(2 votes)
- 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!(3 votes)
- How do I do y+x=3 and -2x+y=6(2 votes)
- Multiply by the first equation by 2 and set them equal or use elimination. From there, it is very straight-forward. You can just solve the equation with one variable.(1 vote)
- how do you solve this problem y=x-2
- Since y = x-2, you can substitute x-2 for y in the second equation.
You now have x-2=4x-2
Solve the equation for x. Take your answer for x and plug it into y=x-2 to get the answer for y.(3 votes)
- How do I answer this 4x-2y=14 and x+2y=6 using substitution method and can you show the proper solving(2 votes)
- whats the answer to 2x+3=y when you have to make x the subject of a formula?(2 votes)
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.