Question

two systems of equations are shown below. the first equation in system b is the original equation in system a. the second equation in system b is the sum of that equation and a multiple of the second equation in system a. a. 1/2 x 3y = 11 → 1/2 x 3y = 11 5x − y = 17 → 15x − 3y = 51 15 1/2 x = 62 b. 1/2 x 3y = 11 15 1/2 x = 62

Answer 1

For Part A: "replacing one equation with the sum of that equation and a multiple of the other" replace first equation by sum of the first equation and some constant times the second equation. After that, replace second equation by sum of the second equation and some constant times the first equation.
You can choose the constant as 2.
Replace (3x + 8y = 12) by (3x + 8y = 12) + 2 * (2x + 2y = 3)
Simplify the above.
(3x+8y=12)+(4x+4y=6)
7x+12y=18
Replace (2x + 2y = 3) by (2x + 2y = 3) + 2 * (3x + 8y = 12)
(2x+2y=3)+(6x+16y=24)
8x+18y=27Part A) The equivalent system of equations are: 7x +12y =18 8x +18y = 27
so the equivalent systems of equations will be 8x+18y=27 7x+12y=18 Now find the value for x and y
Part B) First solve the original system of equations. Then solve the equivalent system of equations created in part A. Prove the solutions for x and y are the same in both cases
Original system of equations:
3x + 8y = 12
2x + 2y = 3
Multiply second equation by -4 and add it to the first. Solve for x.
i got x=0
Sub x = 0 in the second equation and find y
in fractions: (0, 3/2)
Solve the equivalent system created in part A
7x+12y=18
8x+18y=27
-56x-96y=-144
56x+144y=216
48y=72
y=1.5

Answer 2

x = 4, y=3

Explanation:

System A :

`(1)/(2)x+3y=11\\\\5x-y=17`

Since we are given that System b first equation is same as first equation of system a

⇒ System b :

first equation :
`(1)/(2)x+3y=11`

We are also given that second equation in system b is the sum of that equation and a multiple of the second equation in system a.

⇒
`3(5x-y=17)`

⇒
`15x-3y=51`

Now add this equation with first equation of system a.

Thus second equation of system b is

`(15x-3y=51)+((1)/(2)x+3y=11)`

⇒
`_1_5(1)/(2) x =62`

⇒
`(31)/(2)x =62`

⇒
`x=(62*2)/(31)`

⇒
`x=4`

Now to find y put value of x in any equation

we will put in first equation of system a i.e.

`(1)/(2)x+3y=11`

⇒
`(1)/(2)*4+3y=11`

⇒
`2+3y=11`

⇒
`3y=11-2`

⇒
`3y=9`

⇒
`y=(9)/(3)`

⇒
`y=3`

Thus SYSTEM A :

`(1)/(2)x+3y=11\\\\5x-y=17`

SYSTEM B :

`(1)/(2)x+3y=11\\\\(31)/(2)x =62`

Solution of system of equations : x = 4, y=3