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
[tex]\frac{x}{2}-\frac{y}{5}=1[/tex] ----> equation A
[tex]y-\frac{x}{3}=8[/tex] ----> equation B
Multiply equation A by 10 both sides to remove the fractions
[tex]5x-2y=10[/tex] ----> equation C
isolate the variable y in equation B
[tex]y=\frac{x}{3}+8[/tex] ----> equation D
we have the system of equations
[tex]5x-2y=10[/tex] ----> equation C
[tex]y=\frac{x}{3}+8[/tex] ----> equation D
Solve the system by substitution
substitute equation D in equation C
[tex]5x-2(\frac{x}{3}+8)=10[/tex]
solve for x
[tex]5x-\frac{2x}{3}-16=10[/tex]
Multiply by 3 both sides
[tex]15x-2x-48=30[/tex]
[tex]15x-2x=48+30[/tex]
Combine like terms
[tex]13x=78[/tex]
[tex]x=6[/tex]
Find the value of y
[tex]y=\frac{x}{3}+8[/tex]
[tex]y=\frac{6}{3}+8[/tex]
[tex]y=10[/tex]
The solution is the ordered pair (6,10)
Part b) we have
[tex]xy=21[/tex] ---> equation A
[tex]x+5y=22[/tex] ----> equation B
isolate the variable x in the equation B
[tex]x=22-5y[/tex] ----> equation C
substitute equation C in equation A
[tex](22-5y)y=21[/tex]
solve for y
[tex]22y-5y^2=21[/tex]
[tex]5y^2-22y+21=0[/tex]
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
[tex]x=22-5(1.4)=15[/tex]
For y=3
[tex]x=22-5(3)=7[/tex]
therefore
The solutions are the ordered pairs (7,3) and (15,1.4)
