How to solve these simultaneous equations?
Author : az_lender
Submitted : 20180213 23:38:04 Popularity:
Tags: solve simultaneous equations
x7y=2
x^2=34y^2+7xy16
also,
3x+y=5
x^2+2y^2=3x2y
and,
2x3y=3
4x^29y^2=3
Use substitution
1.
x − 7y = 2
x² = 34y² + 7xy − 16
From 1st equation: x = 7y+2
Plug this value of x into 2nd equation and solve for y:
(7y+2)² = 34y² + 7y(7y+2) − 16
49y² + 28y + 4 = 34y² + 49y² + 14y − 16
34y² − 14y − 20 = 0
34y² − 34y + 20y − 20 = 0
34y (y − 1) + 20 (y − 1) = 0
(y − 1) (34y + 20) = 0
y = 1, y = −10/17
x = 7y+2 = 9 or −36/17
Solutions: (9,1) and (−36/17, −10/17)
2.
3x + y = 5
x² + 2y² = 3x − 2y
From 1st equation we get: y = 5−3x
Plug this value of y into 2nd equation and solve for x:
x² + 2(5−3x)² = 3x − 2(5−3x)
x² + 2(25−30x+9x²) = 3x − 2(5−3x)
19x² − 60x + 50 = 9x − 10
19x² − 69x + 60 = 0
x = (69 ± √(69²−4·19·60)) / (2·19) = (69±√201)/38
y = 5−3x = 5−3(69±√201)/38 = (−17∓3√201)/38
Solutions: ((69+√201)/38, (−17−3√201)/38) and ((69−√201)/38, (−17+3√201)/38)
3.
2x − 3y = 3 ..... [ 1 ]
4x² − 9y² = 3
(2x − 3y) (2x + 3y) = 3
3 (2x + 3y) = 3
2x + 3y = 1 ..... [ 2 ]
Add equations [ 1 ] and [ 2 ]
4x = 4
x = 1
Subtract equation [ 1 ] from equation [ 2 ]
6y = −2
y = −1/3
Solution: (1, −1/3)
1).
x  7y = 2
x^2 = 34y^2 + 7xy  16
Solutions:
x = 36/17, y = 10/17
x = 9, y = 1
2).
3x + y = 5
x^2 + 2y^2 = 3x  2y
Solutions:
x = 69/38  sqrt(201)/38, y = (3 sqrt(201))/38  17/38
x = 69/38 + sqrt(201)/38, y = 17/38  (3 sqrt(201))/38
3).
2x  3y = 3
4x^2  9y^2 = 3
Solution:
x = 1, y = 1/3
All systems have the equation of a straight line and a curve, the solution to these systems are the coordinates of the intersection points(if any). Below are the steps to solve:
1) Using the equation of straight line Find x in terms of y or y in terms of x
2) Substituted the x or y from above in the 2nd equation(equation of curve)
3) Solve the resulting quadratic equation
4) Plug in the solution of the quadratic in your first equation to find the coordinated of the other term.
Done.
First one
x = 7y +2
Substitute that into the next equation.
(7y+2)^2 = 34y^2 + 7y(7y+2) 16
49y^2 +28y +4 = 34y^2 +49y^2 +14y  16
0 = 34y^2 14y  20
0 = 17y^2  7y  10
(17y +10)(y  1) =0
y = 10/17 ; 1
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