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saul85 [17]
3 years ago
11

I need help solving a system of linear equation, and it says to solve it by Substitution.

Mathematics
2 answers:
Nata [24]3 years ago
7 0
First solve for x in the second equation (x=-6y+18).then substitute -6y+18 in the first equation and solve for y.after solving for y then substitute it in one of the equation and then solve for x.
2x-3y=-25
x+6y=18

x=-6y+18
2(-6y+18)-3y=-24
-12y+36-3y=-24
-15y+36=-24
-15y=-24-36
-15y=-60
y=4

x+6(4)=18
x+24=18
x=18-24
x=-6

y=4,x=-6
olga_2 [115]3 years ago
6 0
2x - 3y = -24
x+ 6y = 18

you solve the first one by the x:

2x = -24 + 3y

x = (-24+3y)/2

then you substitute what you found in the other equation and you solve by y

(-24+3y)/2 + 6y = 18                   multiply everything by 2

-24 + 3y + 12y = 36

3y + 12y = 36 + 24

15y = 60

y = 60/15

y = 4

Good job! You found your y! :D

Now let's substitute the y in the first equation with 4:

2x - 3(4) = -24

2x - 12 = -24

2x = -24 + 12

2x = -12

x = -12/2

x= -6

Done!! ;) <span />
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Read 2 more answers
AX + B = C what is X equal to?
elena-14-01-66 [18.8K]

Answer:

x is a variable

Step-by-step explanation:

what happens an squared plus B equals C and what I want to do is solve for a so again what we want to do is when we're taking solver a we want to isolate the variable get the variable by itself so the data that need to look at well what is happening on my variable a well you can see me as being multiplied by X and it's being added by B so I need to undo those but we got to make sure we undo them in a certain upper certain order which we call the reverse order of operations which is like the order of operations but the reverse method meaning I'm gonna undo addition or subtraction first so you can see that since my variable a is being added by B I need to undo that by subtracting B and I'll use my subtraction property of equality that's going to now subtract a 0 and then these C minus B are not like terms so I'm going to write ax is equal to C minus B now I need to solve for a so I need to look at and say alright my a is being x over X so the inverse operation of multiplying is dividing by X so therefore have a equals C minus B divided by X now sometimes you could say alright that's correct but we could also divide this X into both of these terms and I'm going to rewrite this in a different form I could say a equals C over X minus B over X alright so what I'm doing is are just dividing those through

hope this helps

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