Question

Pls solve the simultaneous equation in the attachment.

Answer 1

Answer:

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

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

Step-by-step explanation:

Part a) we have

$\frac{x}{2}-\frac{y}{5}=1$ ----> equation A

$y-\frac{x}{3}=8$ ----> 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

$y=\frac{x}{3}+8$ ----> equation D

we have the system of equations

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

$y=\frac{x}{3}+8$ ----> equation D

Solve the system by substitution

substitute equation D in equation C

$5x-2(\frac{x}{3}+8)=10$

solve for x

$5x-\frac{2x}{3}-16=10$

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

$y=\frac{x}{3}+8$

$y=\frac{6}{3}+8$

$y=10$

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)