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Topic: Solve using the substitution method?**Question:**
These are two problems.
1) x+y=6
x-y=4
2) 3x+4y=20
3x-2y=8

June 16, 2019 / By Christabelle

Solve for Y on the bottom one, circle it and draw an arrow to the Y on the top one. Then substitute it where there was a Y. You now only have to solve for one variable just combine like terms and solve for X. You've now found both X & Y

👍 150 | 👎 1

Did you like the answer? First, solve for y. You can use either equation, but the second one is easier (just add 7x to both sides to get y = 7x + 40) Then, whenever you see y, substitute 7x + 40. So You have 4x + 3(7x + 40) = -35 4x + 21x + 120 = -35 25x + 120 = -35 25x = -155 x = -6.25 Then, substitute to get y. y = 7(-6.25) + 40 y = -43.75 + 40 y = -3.75 I hope this information was very helpful.

You need to get a variable (letter) on one side by itself. So in problem 1. you have x + y = 6. So subtract y from both sides and you get x = (6-y) Now plug that value in to the second equation. x- y = 4 becomes (6-y)-y = 4. Or 6-2y = 4 Subtract 6 from both sides. -2y = -2 Divide both sides by -2 and you get y = 1. Since you know that x = (6-y) you substitute the 1 for y. x = 6-1 or 5. So x is 5 and y is 1. For problem 2. 3x + 4y = 20 Subtract 4y 3x = 20 - 4y Divide by 3. x = (20 - 4y)/3 Plug into equation 2. 3x -2y = 8 becomes 3(20-4y)/3 -2y = 8 or (20-4y) -2y = 8 or 20 - 6y =8 Subtract 20 -6y = 8-20 or -12 Divide by -6 y = 2 Since x = (20 -4y)/3 = (20 -8)/3 = 12/3 =4 y is 2 and x is 4.

👍 60 | 👎 -6

1. First equation gives y = 6-x, substitute into second: x - (6-x) = 4 => x - 6 + x = 4 => 2x = 10 => x = 5 y = 6-x = 1. So x = 5 and y = 1. 2. Second equation gives 2y = 3x - 8, substitute into first equation: 3x + 2(3x - 8) = 20 => 3x + 6x - 16 = 20 => 9x = 36 => x = 4 2y = 12 - 8 = 4 => y = 2. So x = 4 and y = 2.

👍 60 | 👎 -13

1) x-y = 4 => x = y+4 sub in y+4 for x in the first equation (y+4) + y = 6 2y= 2 y =1 plug in y = 1 into either equations x +1 = 6 x = 5 y = 1 2) as you see there is the term "3x" in both equations so solve for that in one of them 3x - 2y = 8 => 3x = 2y + 8 sub in 2y + 8 into the first equation for 3x (2y + 8) +4y = 20 6y = 12 y = 2 plug y = 2 into either equation 3x + 8 = 20 3x = 12 x = 4 y = 2

👍 60 | 👎 -20

x+y=6 -------->1 x-y=4 --------->2 add 1&2 =>x+y+x-y=6+4 =>2x=10 =>x=5 substitute x=5 in any of above equations x+y=6 =>5+y=6 =>y=1 therefore solutions are x=5,y=1 3x+4y=20 -------->1 3x-2y=8 --------->2 sub 2 from 1 =>3x+4y-3x+2y=20-8 =>6y=12 =>y=2 substitute y=2 in any of above equations 3x+4y=20 =>3x+8=20 =>x=4 therefore solutions are x=4,y=2

👍 60 | 👎 -27

You must have two equations to solve for two unknown variables. The two unknown variables in this case are the number of one bedroom apartments and the number of two bedroom apartments. One bedroom apartments we'll call "x" Two bedroom apartments we'll call "y" Using the word problem we know a few key things: There are 80 apartments total The cost of each type of apartment The total rent Let's show that we know that the number of apartments is 80: x+y=80 Mathematically that means when you add the number of one and two bedroom apartments you get the total We need another equation: Let's show that we know how to find the total rent using the two types of apartments: 850x+1050y=78400 Now we have two equations with x and y in them To substitute one equation into the other, you must solve one for x or y and put it into the other. The first equation is easier, so as an example let's solve that for x (you could do it for y if you wanted). x+y=80 x=80-y Alright, let's plug the value for x into the other equation 850(80-y)+1050y=78400 Distribute 850 68000-850y+1050y=78400 Simplify 68000+200y=78400 200y=10400 Solve for y y=10400/200 y=52 Now that you know what y is (number of 2 bedroom apartments) plug it into one of the equations to find x. We'll use the first equation because it is simpler. x+52=80 solve for x 80-52=28

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