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. 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

this impossible help me

`Ok, let's take the same problem and break it down, very carefully. *2y = x + 7 & x = y - 4* so 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 - 4* *x = _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 :)`

2x+3y=10 3x+4y=8 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

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.

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

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!

Main content

8th grade

Course: 8th grade > Unit 4

Lesson 3: Solving systems with substitution

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

Name: Systems of equations with substitution: 2y=x+7 & x=y-4
Uploaded: 2011-03-16T18:27:52Z
Description: 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.

Google Classroom

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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wsspjs
Posted 10 years ago. Direct link to wsspjs's post “Trying to solve substitut...”
Trying to solve substitution problem.
x=3y-4
4y-2x=13
Do I need to use a negative fraction with X?
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(11 votes)
Answer
- Sid
  Posted 10 years ago. Direct link to Sid's post “Just substitute the value...”
  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.
  Comment on Sid's post “Just substitute the value...”
  (10 votes)
Lori Steyer Hoag
Posted 10 years ago. Direct link to Lori Steyer Hoag's post “How do I solve an equatio...”
How do I solve an equation like this? X=y+8. And 3x-y=16 ? I am trying to solve for x and y
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(7 votes)
Answer
- Chamuel Patino
  Posted 10 years ago. Direct link to Chamuel Patino's post “x=y+8. Substitute that in...”
  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
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  (9 votes)
yasgup332
Posted 8 months ago. Direct link to yasgup332's post “this impossible help me”
this impossible help me
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(2 votes)
Answer
- NEL_MAN31
  Posted 6 months ago. Direct link to NEL_MAN31's post “`Ok, let's take the same ...”
  Ok, let's take the same problem and break it down, very carefully. 2y = x + 7 & x = y - 4 so 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 - 4 x = 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 :)
  Comment on NEL_MAN31's post “`Ok, let's take the same ...”
  (19 votes)
JJ
Posted 7 months ago. Direct link to JJ's post “Can't we just put it into...”
Can't we just put it into Desmos caculator and solve it that way?
Button navigates to signup pageComment on JJ's post “Can't we just put it into...”
(4 votes)
Answer
SOPHIAM
Posted 8 months ago. Direct link to SOPHIAM's post “how do i watch the video ...”
how do i watch the video again its not letting me
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(3 votes)
Answer
Cassidy Smith
Posted 10 years ago. Direct link to Cassidy Smith's post “2x+3y=10 3x+4y=8 How do ...”
2x+3y=10
3x+4y=8

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
Button navigates to signup pageComment on Cassidy Smith's post “2x+3y=10 3x+4y=8 How do ...”
(2 votes)
Answer
- Mr. Jones
  Posted 10 years ago. Direct link to Mr. Jones's post “You must first solve one ...”
  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.
  Button navigates to signup page
  (3 votes)
lrf73095
Posted 9 years ago. Direct link to lrf73095's post “How do i solve 5y(2y-3)+(...”
How do i solve 5y(2y-3)+(2y-3)?
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(2 votes)
Answer
- Anu
  Posted 9 years ago. Direct link to Anu's post “5y(2y-3) + (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
  Button navigates to signup page
  (3 votes)
maryam.aimaq
Posted 8 months ago. Direct link to maryam.aimaq's post “-6x+6y=-24 -3x-12y=-12”
-6x+6y=-24
-3x-12y=-12
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(2 votes)
Answer
- Ottomatic
  Posted 5 months ago. Direct link to Ottomatic's post “y=0 x=4 is one way to sol...”
  y=0
  x=4
  is one way to solve that problem
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  (2 votes)
andy
Posted 3 years ago. Direct link to andy's post “Why does substitution wor...”
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!
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(2 votes)
Answer
samantha.musasa
Posted 6 years ago. Direct link to samantha.musasa's post “how do I solve y=x^2-5x-1...”
how do I solve y=x^2-5x-14 and y=x-7
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(1 vote)
Answer
- Ian Pulizzotto
  Posted 6 years ago. Direct link to Ian Pulizzotto's post “Since both equations give...”
  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!
  Comment on Ian Pulizzotto's post “Since both equations give...”
  (3 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.