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
----> equation A
----> equation B
Multiply equation A by 10 both sides to remove the fractions
----> equation C
isolate the variable y in equation B
----> equation D
we have the system of equations
----> equation C
----> equation D
Solve the system by substitution
substitute equation D in equation C
![5x-2(\frac{x}{3}+8)=10](https://tex.z-dn.net/?f=5x-2%28%5Cfrac%7Bx%7D%7B3%7D%2B8%29%3D10)
solve for x
![5x-\frac{2x}{3}-16=10](https://tex.z-dn.net/?f=5x-%5Cfrac%7B2x%7D%7B3%7D-16%3D10)
Multiply by 3 both sides
![15x-2x-48=30](https://tex.z-dn.net/?f=15x-2x-48%3D30)
![15x-2x=48+30](https://tex.z-dn.net/?f=15x-2x%3D48%2B30)
Combine like terms
![13x=78](https://tex.z-dn.net/?f=13x%3D78)
![x=6](https://tex.z-dn.net/?f=x%3D6)
<em>Find the value of y</em>
![y=\frac{6}{3}+8](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B6%7D%7B3%7D%2B8)
![y=10](https://tex.z-dn.net/?f=y%3D10)
The solution is the ordered pair (6,10)
Part b) we have
---> equation A
----> equation B
isolate the variable x in the equation B
----> equation C
substitute equation C in equation A
![(22-5y)y=21](https://tex.z-dn.net/?f=%2822-5y%29y%3D21)
solve for y
![22y-5y^2=21](https://tex.z-dn.net/?f=22y-5y%5E2%3D21)
![5y^2-22y+21=0](https://tex.z-dn.net/?f=5y%5E2-22y%2B21%3D0)
Solve the quadratic equation by graphing
The solutions are y=1.4, y=3
see the attached figure
<em>Find the values of x</em>
For y=1.4
![x=22-5(1.4)=15](https://tex.z-dn.net/?f=x%3D22-5%281.4%29%3D15)
For y=3
![x=22-5(3)=7](https://tex.z-dn.net/?f=x%3D22-5%283%29%3D7)
therefore
The solutions are the ordered pairs (7,3) and (15,1.4)