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## Class 10 math (India)

### Course: Class 10 math (India)>Unit 3

Lesson 2: Algebraic methods

# Systems of equations with substitution: -3x-4y=-2 & y=2x-5

Learn to solve the system of equations -3x - 4y = -2 and y = 2x - 5 using substitution. Created by Sal Khan.

## Want to join the conversation?

• In a former video, an evil advice-giver dude was humiliated by a smart talking bird. How Sal came up with that, I have no freaking clue.
• i get how to solve for y but how do you solve for x
• just plug in the value of y into one of the original equations and solve for x
• Every time there is a video about systems of equations with substitution there is always 2 equations. Could you find the values of x and y (if you had 2 unknown variables) with only one equation?
• If you have one equation with 2 variables (or a linear equation like 2x + 5y = 20), there are an infinite set of solutions. This type of equation creates a line where each point on the line represents an (x, y) ordered pair that is a solution to the equation.

When you have 2 equations with the same 2 variables, then you have a system of linear equations. The solution to the system is the point (or points) that the 2 linear equations have in common.

Hope this helps.
• I am experiencing brain fog. I have a test tmrw. Any advice??
• It’s okay to study the night before the test, but don’t stay up too late studying. It is best to get a good night’s sleep before the test.
• …. um..…. question mark?
• In a former video, a king's advisor was humiliated by a smart talking bird who can do math.
2x-8=y and 3x-2y=0
y=x+6 and y=2x-7
2x-y=3 and y=x+4?
• For the first one, y = 4x -3 and y = 3x -3, you can substitute "-3x-3" in for y. Now you have:

-3x-3 = 4x - 3

Move the x's to one side and the numbers to the other to get:
7x = 0. So, x is 0.

Plug that back in and you find that y = -3.

use the same principle for the others
• what is 2x=16-8y but you have to substitute x+4y=25 how would you do this
• To substitute, you have two choices to isolate variables, in both equations, solving for x is the easiest. In the first equation, you could divide by 2 to get x=8-4y. If you have 8-4y+4y=25, you end up with 8=25, so there is no solution (lines parallel).
If you subtract 4y in second equation, you get x=25-4y and substituting in first gives 2(25-4y)=16-8y, distribute to get 50-8y=16-8y, so when you add 8y to both sides, 50=16 which also gives no solution.
This can be seen by getting both in slope intercept form:
y=-1/4 x + 2 and y=-1/4 x + 25/4, both have same slope and different intercepts.
• What do you do if there is a third variable and you have to solve for it?
I'm particularly having trouble with this equation with finding 'm'.

3x + my = 5
(m+2)x + 5y = m

In addition i have to find 'm' when there are infinite solutions and no solution.

sorry for the long question but i'm just not understanding this, would mean a lot if someone helped me out here.
• 3𝑥 + 𝑚𝑦 = 5 ⇒ 𝑦 = −(3∕𝑚)𝑥 + 5∕𝑚

(𝑚 + 2)𝑥 + 5𝑦 = 𝑚 ⇒ 𝑦 = −((𝑚 + 2)∕5)𝑥 + 𝑚∕5

Viewing 𝑚 as a constant, each of the two equations describe a straight line.

For the system to have infinitely many solutions, both lines must have the same slope AND the same 𝑦-intercept.

For the system to have no solutions, the two lines must have the same slope, but different 𝑦-intercepts.

So, first of all we want to know when the two lines have the same slope, which means we want to solve the equation
−3∕𝑚 = −(𝑚 + 2)∕5

Multiplying both sides by (−5𝑚) we get
15 = 𝑚(𝑚 + 2)

Distributing the 𝑚 and subtracting 15 from both sides we get
𝑚² + 2𝑚 − 15 = 0, which has the two solutions
𝑚 = −5, 𝑚 = 3

Next, we want to know when the 𝑦-intercepts are equal:
5∕𝑚 = 𝑚∕5

Multiplying both sides by 5𝑚 we get
𝑚² = 25 ⇒ 𝑚 = ±5

So, if the system has infinitely many solutions, then 𝑚 = −5,
and if the system doesn't have any solutions, then 𝑚 = 3
• What would 250m = pc be? (p and c are both variables)
• what about something like this:

43x+6y=87
20x-2y=74

I need to find what x and what y is. I am stuck. Can anyone explain how to do these problems please?
Thanks
• 43x+6y=87
43x=−6y+87
43x/43 = -6y+87/43
x=-6/43y + 87/43

Subsitute -6/43y + 87/43 for x in 20x-2y=74
20x-2y=74
20(-6/43y+87/43)-2y=74
-206/43 y+1740/43 = 74 ( simplify both sides of the equation)
-206/43 y +1740/43 + -1740/43 =74=-1740/43 (add(-1740)/43 to both sides)
-206/43 y = 1442/43 ( divide both sides)
y=-7
Subsitute
x=-6/43 y+87/43
x=-6/43(-7)+87/43 ( simplify)
x=3
x=3 and y=-7