Question

Pls solve the simultaneous equation in the attachment.

Answer 1

Part a) The solution is the ordered pair (6,10)

Part b) The solutions are the ordered pairs (7,3) and (15,1.4)

Explanation:

Part a) we have

`(x)/(2)-(y)/(5)=1` ----> equation A

`y-(x)/(3)=8` ----> equation B

Multiply equation A by 10 both sides to remove the fractions

`5x-2y=10` ----> equation C

isolate the variable y in equation B

`y=(x)/(3)+8` ----> equation D

we have the system of equations

`5x-2y=10` ----> equation C

`y=(x)/(3)+8` ----> equation D

Solve the system by substitution

substitute equation D in equation C

`5x-2((x)/(3)+8)=10`

solve for x

`5x-(2x)/(3)-16=10`

Multiply by 3 both sides

`15x-2x-48=30`

`15x-2x=48+30`

Combine like terms

`13x=78`

`x=6`

Find the value of y

`y=(x)/(3)+8`

`y=(6)/(3)+8`

`y=10`

The solution is the ordered pair (6,10)

Part b) we have

`xy=21` ---> equation A

`x+5y=22` ----> equation B

isolate the variable x in the equation B

`x=22-5y` ----> equation C

substitute equation C in equation A

`(22-5y)y=21`

solve for y

`22y-5y^2=21`

`5y^2-22y+21=0`

Solve the quadratic equation by graphing

The solutions are y=1.4, y=3

see the attached figure

Find the values of x

For y=1.4

`x=22-5(1.4)=15`

For y=3

`x=22-5(3)=7`

therefore

The solutions are the ordered pairs (7,3) and (15,1.4)