Question

Select the correct answer from the drop-down menu.

Answer 1

Answer:

2. $(-\frac{1}{2},\frac{13}{2})$

Step-by-step explanation:

Given Equations:

$9x + 5y = 28$          Equation:1

$5x + 9y = 56$           Equation:2

Multiplying Equation:1 by 5 and Equation:2 by 9, gives

$45x+25y=140$    Equation:3

$45x+81y=504$     Equation:4

Using Elimination Method, Subtract Equation:3 from Equation:4

$45x+81y-(45x+25y)=504-140$

$45x+81y-45x-25y=504-140$

$56y=364$

$y=\frac{364}{56}\\\\ y=6.5$  or $y=\frac{13}{2}$

Putting, the value of 'y' in Equation:1

$9x + 5y = 28\\9x + 5(6.5) = 28\\9x+32.5=28\\9x=28-32.5\\9x=-4.5\\x=-\frac{4.5}{9} \\x=-0.5$

Or , $x=-\frac{1}{2}$

The solution for the set of equations is: $(-\frac{1}{2},\frac{13}{2})$

Answer 2

Answer:

The answer to your question is 2. (-1/2, 13/2)

Step-by-step explanation:

Equation 1      9x + 5y = 28

Equation 2     5x + 9y = 56

Process

- Solve by Elimination

- Multiply equation 1 by 5 and equation 2 by -9

                   5(9x + 5y = 28)

                  -9(5x + 9y = 56)

           

                   45x + 25y = 140

                  -45x - 81y   = -504

                     0     -56y = -364

                                  y = -364/-56

                                 y = 13/2 = 6.5

- Find x

                     9x + 5(13/2) = 28

                     9x + 65/2 = 28

                     9x = 28 - 65/2

                     9x = -9/2

                       x = (-9/2)/9

                       x = -1/2