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frozen [14]
4 years ago
14

Question 1. In the below system, solve for y in the first equation.

Mathematics
1 answer:
Phantasy [73]4 years ago
7 0

Answer for Question (1):

The system of equations: x+3y=6

2x-y=10

Solve the first equation x+3y=6 for y:

Subtracting x on both sides,

x+3y-x=6-x

3y=6-x

Now dividing 3 on both sides, we get

y=2-x/3 = negative one third x + 2

Thus y= negative one third of x +2

Answer for question (2):

The system of equation :

5x-3y=-3 --> (1)

2x-6y=-6 -->(2)

Multiply equation (1) by 2, we get 10x-6y=6.

Subtracting (1) by (2), we get 8x=0 implies x=0.

Plug in x=0 in equation (1), 0-3y=-3

Dividing -3 on both sides,

y=-3/-3=1

So the solution of y is 1.

Answer for Question (3):

System of equation:

7x-2y=-3 (1)

14x+y=14 (2)

Multiply (2) by 2,

28x+2y=28 (3)

Adding (1) and (3), we get 35x=25

Now dividing 35 on both sides, x=25/35=5/7.

Plug in x=5/7 in equation (1), we get 5-2y=-3

-2y=-3-5

-2y=-8

Dividing -2 on both sides,

y=4

Thus the solution of this system of equation as (5/7,4).

Answer for question (4):

x + 3y = 9 ---> (1)

3x − 3y = −13 ---> (2)

Adding equation (1) and (2), we get

4x=-4

Dividing 4 on both sides,

x=-1

Plug in x=-1 in equation (1), we get

-1+3y=9

Adding 1 on both sides,

3y=9+1=10

Now dividing 3 on both sides,

y=10/3.

so the solution is (-1,10/3).

Answer for question (5):

System of equation is 4x − y = 3 (1)

7x − 9y = −2 (2)

consider the first equation 4x-y=3

Adding y on both sides, 4x=3+y

subtracting 3 on both sides, we get y=4x-3

Substitute y=4x-3 in equation (2),

7x-9(4x-3)=-2

7x-36x+27=-2

Combine the like terms,

-29x+27=2

Adding 27 on both sides,

-29x=-29

dividing -29 on both sides,

x=1

Plug in x=1 in y=4x-3,

y=4(1)-3=1

Then the solution is (1,1).

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o-na [289]

Answer:

I dunno understand if what kind of opperation you're doing. Is there a multiplication?

<h3>Step-by-step explanation:</h3>

0.60 · 10,000 % = 0.00006 % · 0.6<em>x </em>· 10-5%

If it's a multiplication the answer will be

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3 0
3 years ago
Jordans have a rectangular garden. want to build a diagonal pathway across he garden. the length of pathway is 90 ft. width of g
blondinia [14]
So we see the pythagorean theorem
a^2+b^2=c^2
diagonal=c
width=a or b, pick one
width=a
legnth=b
width is 5 times more than 2 times legnth of garden
w=5(2legnth)
this doesn't make sense since the legnth is normally longer than the width, but we'll stick with that
w=5(2l)
w=10l
a=10b
(10b)^2+b^2=90^2
100b^2+b^2=8100
101b^2=8100
divide by 101
b^2=80.198 aprox 80.2
square root
8.95533
legnth of garden =8.96
to find width subsitue
w=10l
w=89.56

legnth=8.96
width=89.56
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Need help with this question thanks :)
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