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 Your answer is (9, 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

To extend this concept, can you solve a *linear system of equations* with *3 unknowns*? ``` x + y + z = 4 x - 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 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)

Main content

8th grade (Eureka Math/EngageNY)

Course: 8th grade (Eureka Math/EngageNY) > Unit 4

Lesson 4: Topic D: Systems of linear equations and their solutions

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

Name: Systems of equations with substitution: y=-5x+8 & 10x+2y=-2
Uploaded: 2011-03-16T18:27:52Z
Description: Learn to solve the system of equations y = -5x + 8 and 10x + 2y = -2 using substitution.

CCSS.Math:

8.EE.C.8

8.EE.C.8b

Google Classroom

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.

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fazal husain
Posted 10 years ago. Direct link to fazal husain's post “What should u do if they ...”
What should u do if they ask u to verify it without a graph ?
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(4 votes)
Answer
Luzlil Villalobos
Posted 9 years ago. Direct link to Luzlil Villalobos's post “How would I be able to so...”
How would I be able to solve the variable for x and y I'm the following equation
x+3y=12
x-y=8
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(4 votes)
Answer
- Theresa Johnson
  Posted 9 years ago. Direct link to Theresa Johnson's post “To solve a system like th...”
  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
  
  Your answer is (9, 1)
  Comment on Theresa Johnson's post “To solve a system like th...”
  (7 votes)
thabani1813
Posted 7 years ago. Direct link to thabani1813's post “Solve the set of linear e...”
Solve the set of linear equations :
3x-4y=1
4x -3y=6
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(3 votes)
Answer
Sarah Richer
Posted 10 years ago. Direct link to Sarah Richer's post “How would you solve this ...”
How would you solve this problem?
4x + 5y = 11
y = 3x - 13
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(2 votes)
Answer
- Bill
  Posted 10 years ago. Direct link to Bill's post “Once you find x, you will...”
  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)
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  (2 votes)
Luzlil Villalobos
Posted 9 years ago. Direct link to Luzlil Villalobos's post “How would I be able to so...”
How would I be able to solve the variable x and y on the following equation

4x-3y=12
x+2y=14
Button navigates to signup pageComment on Luzlil Villalobos's post “How would I be able to so...”
(3 votes)
Answer
- Chuck Towle
  Posted 9 years ago. Direct link to Chuck Towle's post “Luzlilvillalobos, 4x-3y=1...”
  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
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  (4 votes)
Pei-Hsin Lin 林培心 🌸
Posted 6 years ago. Direct link to Pei-Hsin Lin 林培心 🌸 's post “To extend this concept, c...”
To extend this concept, can you solve a linear system of equations with 3 unknowns?
x + y + z = 4 x - 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.
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(3 votes)
Answer
- Patrick
  Posted 6 years ago. Direct link to Patrick's post “x + y + z = 4 (1) x - y +...”
  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).
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  (4 votes)
dogtag0611
Posted 9 years ago. Direct link to dogtag0611's post “How would solves equation...”
How would solves equations with squares? For example:

xy=12
x^2+y^2=40
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(3 votes)
Answer
Robin Friend
Posted 9 years ago. Direct link to Robin Friend's post “How do I solve y=2x+7 and...”
How do I solve y=2x+7 and y=9x+8 using substitution?
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(3 votes)
Answer
jacka36
Posted 9 years ago. Direct link to jacka36's post “solve for x and y by subs...”
solve for x and y by substitution method
x+y= -1
x^2+y^2=13
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(2 votes)
Answer
- Allen Anderson
  Posted 9 years ago. Direct link to Allen Anderson's post “First, solve the first eq...”
  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
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  (1 vote)
bilalyounus33
Posted 6 years ago. Direct link to bilalyounus33's post “if there is two equation ...”
if there is two equation and ask about the value of 3rd equation how to solve that
like if 19X+17y=5 and 17x+13y=1 then find the value of 91x+47y ??
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(2 votes)
Answer
- Jaime Cheung
  Posted 6 years ago. Direct link to Jaime Cheung's post “You can solve the equatio...”
  You can solve the equations first by minus the second equations by the first one. And when you get the solution put the solutions in the 3rd equation.
  Comment on Jaime Cheung's post “You can solve the equatio...”
  (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.