Question

Due now pls help me!!!!!!!!!!!!!!!!!!

Answer 1

D (1,4)

Explanation:

Given : 5x + 6y = 29 and 3x + 6y = 27

We can solve this system using the elimination method.

What is the elimination method?

The elimination method can be used when the two equations have variables with the same number before. E.g. both equations have 6y. When this happens, you can subtract the two equations resulting in the variables cancelling out. Once, the variables cancel out, you are able to solve for the other variable as they won't cancel out.

Subtracting equation 2 from equation 1

5x + 6y = 29

- 3x + 6y = 27

---------------------

2x + 0y = 2

Solving for x

==> 2x = 2

==> x = 1

Finding y

Now that we have the value of one of the variables, we can plug it into one of the equations and solve for the other variable.

==>5x + 6y = 29 x = 1

==> 5(1) + 6y = 29 multiply 5 and 1

==> 5 + 6y = 29 subtract 5 from both sides

==> 6y = 24 divide both sides by 6

==> y = 4

In the solution, y = 4 and x = 1, meaning that the solution to the system is (1,4)

Answer 2

Answer: D (1,4)

We have the system of equations:

`\boldsymbol{\sf{5x+6y=29 }}\\ \\ \boldsymbol{\sf{3x+6y=27 }}`

From equation 1 we express and

`\boldsymbol{\sf{5x+6y=29 }}`

We pass the addend with the variable x from the left to the right, and we change the sign.

`\boldsymbol{\sf{6y=29-5 }}\\ \\ \boldsymbol{\sf{6y=29-5x}}`

We divide both parts of the equation by the multiplier y

`\boldsymbol{\sf{(6y)/(6)=(29-5x)/(6) \iff \ y=(29)/(6)-(5x)/(6) }}`

We put the result and in equation 2:

`\boldsymbol{\sf{3x+6y=27 }}`

We obtain:

`\boldsymbol{\sf{3x+6\cdot\left((29)/(6)-(5x)/(6)\right)=27 \iff \ 29-2x=27 }}`

We pass the addend, free 29 from the left to the right, changing the 29.

`\boldsymbol{\sf{-2x=-27+29 \iff \ -2x=-2 }}`

We divide both parts of the equation by the multiplier of x.

`\boldsymbol{\sf{((-1)2x)/(-2)=-(2)/(-2) \iff \ x=1 }}`

`\boldsymbol{\sf{What \ y=(29)/(6)-(5x)/(6),then \ y=(29)/(6)-(5)/(6) \iff \ y=4 }}`