Question 3:
Solution:
From (1),y=2x−43 -------(3)Substitute (3) into (2),(2x−43)2+4x2=2(4x2−16x+169)+4x2=24x2−16x+16+36x2=18 (×9)40x2−16x−2=020x2−8x−1=0(10x+1)(2x−1)=0x=−110 or x=12Substitute the values of x into (3),When x=−110,y=2(−110)−43=−125When x=12,y=2(12)−43=−33=1The solutions are x=−110, y=−125 and x=12, y=1.
Solve the following simultaneous equations.
3y – 2x= – 4
y2 + 4x2 = 2
Solution:
3y – 2x= – 4 -----(1)
y2 + 4x2= 2 -----(2)
From (1),y=2x−43 -------(3)Substitute (3) into (2),(2x−43)2+4x2=2(4x2−16x+169)+4x2=24x2−16x+16+36x2=18 (×9)40x2−16x−2=020x2−8x−1=0(10x+1)(2x−1)=0x=−110 or x=12Substitute the values of x into (3),When x=−110,y=2(−110)−43=−125When x=12,y=2(12)−43=−33=1The solutions are x=−110, y=−125 and x=12, y=1.
Question 4:
Solution:
a=3, b=−6, c=2y=−b±√b2−4ac2ay=−(−6)±√(−6)2−4(3)(2)2(3)y=6±√126y=1.577 or 0.423
Solve the simultaneous equations x – 3y = –1 and y + yx – 2x = 0.
Give your answers correct to three decimal places.x – 3y = –1 -----(1)
y + yx – 2x = 0 -----(2)
From (1),
x = 3y – 1 -----(3)
Substitute (3) into (2),
y + y (3y – 1) – 2(3y – 1) = 0
y + 3y2 – y – 6y+ 2 = 0
3y2 – 6y + 2 = 0a=3, b=−6, c=2y=−b±√b2−4ac2ay=−(−6)±√(−6)2−4(3)(2)2(3)y=6±√126y=1.577 or 0.423
Substitute the values of y into (3).
When y = 1.577,
x = 3 (1.577) – 1 = 3.731 (correct to 3 decimal places)
When y = 0.423,
x = 3 (0.423) – 1 = 0.269 (correct to 3 decimal places)
The solutions are x = 3.731, y = 1.577 and x = 0.269, y = 0.423.