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Answers fast please. 1.) y=x−63x+2y=8 Use the substitution method. A.) (4, −2) B.) (14, 8) C.) (0, −6) D.) (3, −3) 2.) What is the x-coordinate of the solution for the system of equations? {y−x=910+2x=−2y X= what? 3.) What is the y-coordinate of the solution for the system of equations? {y=3x−1x−y=−9 Y=what?

Respuesta :

1.
A.) (4, −2)

2.
x = -7
y = 2

3.
x = 5
y = 14
kudzordzifrancis

QUESTION 1

The given system of equation is

[tex]y=x-6...eqn1[/tex]

and


[tex]3x+2y=8...eqn2[/tex]


Let us substitute equation (1) into equation (2) to get,

[tex]3x+2(x-6)=8[/tex]


We expand the bracket to get,

[tex]3x+2x-12=8[/tex]


We simplify to get.

[tex]5x-12=8[/tex]


We group like terms to get

[tex]5x=8+12[/tex]

[tex]\Rightarrow 5x=20[/tex]

[tex]\Rightarrow x=4[/tex]


We now substitute [tex]x=4[/tex] in to equation (1) to obtain,


[tex]y=4-6=-2[/tex]


The correct answer is option A.


QUESTION 2

The given system of equations is

[tex]y-x=9...eqn1[/tex]


and

[tex]10+2x=-2y...eqn2[/tex]

We make y the subject in equation (2) to get,

[tex]y=-x-5..eqn3[/tex]


We put equation (3) into equation (1) to obtain,


[tex]-x-5-x=9[/tex]


We group like terms to get,

[tex]-x-x=9+5[/tex]


This implies that,

[tex]-2x=14[/tex]


We divide through by [tex]-2[/tex] to get,

[tex]x=-7[/tex]


Hence the x-coordinate is [tex]-7[/tex]



QUESTION 3

The given system is

[tex]y=3x-1..eqn1[/tex]


and


[tex]x-y=-9...eqn2[/tex]

We make [tex]x[/tex] the subject in equation (2) to get,

[tex]x=y-9...eqn3[/tex]


We put equation (3) into equation (1) to obtain,


[tex]y=3(y-9)-1[/tex]


We expand the bracket to get,


[tex]y=3y-27-1[/tex]


Group like terms to get,


[tex]y-3y=-27-1[/tex]


We simplify to get;

[tex]-2y=-28[/tex]


This implies that,

[tex]y=14[/tex]


Therefore the y-coordinate is 14.

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