Respuesta :

calculista

Keywords

system of equations, linear equation, elimination method

we have

[tex]\frac{1}{3}x+ \frac{1}{4}y=1[/tex]

Multiply by [tex]12[/tex] both sides

[tex]4x+3y=12[/tex] -----> linear equation A

[tex]2x-3y=-30[/tex] -----> linear equation B

Applying the elimination method for solve the system of equations

Adds equation A and equation B

[tex]4x+3y=12 \\2x-3y=-30\\-------- \\4x+2x=12-30 \\6x=-18 \\x=-3[/tex]

Find the value of y

[tex]4(-3)+3y=12[/tex]

[tex]3y=12+12[/tex]

[tex]y=8[/tex]

therefore

the answer is the option D

[tex]8[/tex]

Their is the system of linear equations [tex]\frac{1}{3} x+\frac{1}{4} y=1[/tex] and [tex]2x-3y=-30[/tex].

The solution of the given system of equation is 8. The correct answer is option D.

Given equations:

      [tex]\frac{1}{3} x+\frac{1}{4} y=1[/tex]

      [tex]2x-3y=-30[/tex]

Let

    [tex]\frac{1}{3} x+\frac{1}{4} y=1[/tex]         ...........(i)

    [tex]2x-3y=-30[/tex]     ..........(ii)

Look for terms that can be eliminated. The equations do not have any x or y terms with the same coefficients.

Multiply the equation (i) by 12 so they do have the same coefficient.

   [tex]12(\frac{1}{3} x+\frac{1}{4} y=1)[/tex]

   [tex]4x+3y=12[/tex]       .............(iii)

Adding equation (ii) and equation (iii)

     [tex]2x-3y=-30\\4x+3y=12[/tex]

  ------------------------

    [tex]6x=-18\\x=\frac{-18}{6}\\x=-3[/tex]

Substitute x = -3 into one of the original equations to find y.

   [tex]2(-3)-3y=-30\\-6-3y=-30\\-3y=-30+6\\y=\frac{-24}{-3} \\y=8[/tex]

Therefore, the answer is the option D

For more information:-

https://brainly.com/question/12735283

   

Otras preguntas