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My name is Ann [436]
3 years ago
12

How do you solve system of eqautions using substitution

Mathematics
1 answer:
Licemer1 [7]3 years ago
8 0
Well basically you are plugging in variables. Lets use a random but simple system of equations.
y = 2x - 3
6x + 8y = 20
Now you need to plug in the top equation, which equals y, into the bottom equation.
6x + 8(2x - 3) = 20
6x + 16x - 24 = 20
Combine like terms.
22x - 24 = 20
Add 24 to both sides.
22x = 44
Divide.
x = 2
Now plug in 2 in any of the original equations.
y = 2x - 3
y = 2(2) - 3
y = 4 - 3
y = 1
So your solution is (2,1). And thats how you use substitution.
I hope this helps love! :)
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Answer:

answer is of left to right

16x^14/(36x^2)=4x^12/9

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-9a^3(b^5)

7 0
3 years ago
X-3y=-6 what is the solution for this question ??
seraphim [82]

Answer: y = 1/3x + 2

8 0
3 years ago
Please explain and show work
Vinvika [58]
1*3/5 = 3/5.
3/5(t-6) = -0.4
Distribute 3/5
3/5t - 3.6 = -0.4
Add 3.6 
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Divide by 3/5 
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6 0
3 years ago
Read 2 more answers
How many integers between 100 and 1000 have both 15 and 18 as factors?
Zanzabum

The numbers 180, 270, 360, 450, 540, 630, 720, 810, and 900 are the integers between 100 and 1000 that have both 15 and 18 as a factor.

<h3>What is an integer?</h3>

it is defined as the whole number it can be a negative whole number or a positive whole number for an example: 2, 4, 1, -8, 0, etc.

To find the how many integers between 100 and 1000 have both 15 and 18 as factors.

We must calculate the LCM of 15 and 18

The LCM for 15 and 18 is 90

Now the 90's multiple:

90, 180, 270, 360, 450, 540, 630, 720, 810, 900, 1080

These numbers are factors of 15 and 18 also.

The number which lies between 100 and 1000 are:

180, 270, 360, 450, 540, 630, 720, 810, 900

Thus, the numbers 180, 270, 360, 450, 540, 630, 720, 810, and 900 are the integers between 100 and 1000 that have both 15 and 18 as a factor.

Learn more about the integers here:

here:brainly.com/question/15276410

#SPJ1

4 0
2 years ago
Solve for c: a(c−b)=d
Sindrei [870]

Answer:

a(c-b) = d

Use the distributive property, which states: x(y-z) = xy - xz

a(c-b) = ac - ab

ac - ab = d

Add ab on both sides

ac = d + ab

Divide both sides by a to isolate c

(If you would, please mark branliest, I'm one away from getting Expert

7 0
3 years ago
Read 2 more answers
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