Answer:
WHEN x=4 , y=[tex]\frac{1}{2}[/tex] or 0.5 (4,0.5)
WHEN x=[tex]\frac{1}{3}[/tex] , y=6 ([tex]\frac{1}{3}[/tex],6)
Step-by-step explanation:
equation (1) 3x+2y=13
equation(2) xy=2
step1: make x the subject of equation (1).
3x+2y=13
(subtract 2y for both sides)
3x=13-2y
(divide by 3 for both sides)
x=[tex]\frac{13-2y}{3}[/tex]
step2: sub the value of x into the second equation.
xy=2
([tex]\frac{13-2y}{3}[/tex])y=2
(expand the bracket)
[tex]\frac{y(13-2y)}{3}[/tex]=2
(multiply by 3 for both sides)
y(13-2y)=6
(expand the bracket)
13y-2[tex]y^{2}[/tex]=6
(subtract 6 for both sides)
13y-2[tex]y^{2}[/tex]-6=0
(factorise the expression)
(-2y+1)(y-6)=0
step3: solve for y.
1. -2y+1=0
-2y=-1
y=[tex]\frac{1}{2}[/tex] (the answer)
2. y-6=0
y=6 (the answer)
step4: sub the values of y into one of the equations and solve for x.
1.xy=2
x(0.5)=2
(divide by 0.5 for both sides)
x=4 (the answer)
2. xy=2
x(6)=2
(divide by 6 for both sides)
x=[tex]\frac{1}{3}[/tex] (the answer)
