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### Course: 8th grade > Unit 4

Lesson 3: Solving systems with substitution- Systems of equations with substitution: 2y=x+7 & x=y-4
- Systems of equations with substitution
- Systems of equations with substitution: y=4x-17.5 & y+2x=6.5
- Systems of equations with substitution: -3x-4y=-2 & y=2x-5
- Systems of equations with substitution: 9x+3y=15 & y-x=5
- Systems of equations with substitution
- Systems of equations with substitution: y=-5x+8 & 10x+2y=-2
- Systems of equations with substitution: y=-1/4x+100 & y=-1/4x+120
- Substitution method review (systems of equations)

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

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.

## Want to join the conversation?

- What should u do if they ask u to verify it without a graph ?(6 votes)
- Put both equations in slope y-intercept form

If both have the same slope and different

y-intercept then these lines are parallel

and the system has no solution(4 votes)

- 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**.(6 votes)- 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)*.(7 votes)

- How would I be able to solve the variable for x and y I'm the following equation

x+3y=12

x-y=8(4 votes)- 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)(7 votes)

- Solve the set of linear equations :

3x-4y=1

4x -3y=6(3 votes) - How would you solve this problem?

4x + 5y = 11

y = 3x - 13(2 votes)- 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)(3 votes)

- How would I be able to solve the variable x and y on the following equation

4x-3y=12

x+2y=14(3 votes)- 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(4 votes)

- How would solves equations with squares? For example:

xy=12

x^2+y^2=40(3 votes) - So if two equations have the same slope then they are parallel and will never intersect? Could you therefore refer to the value on the X coefficients as proof without substitution that a system of equations in slope intersect form has no solution? Like a shortcut.(2 votes)
- Yes, if you can write each equation in slope-intercept form and show that the coefficients on x are the same while the constant terms are different, this would prove that the lines never intersect (no solution).

Have a blessed, wonderful day!(2 votes)

- How would i solve this problem? The square root of 13z+2x(2 votes)
- you would not. If you have a variable with no square symbol, you can't square it unless you know what z and x are.(2 votes)

- How do I solve y=2x+7 and y=9x+8 using substitution?(3 votes)

## 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.