Download - Simultaneous equations
K S Sandhu
Simultaneous equationsSolve y = 2x2 – 3x + 4 and y = 4x + 1
2x2 – 3x + 4 = 4x + 1
2x2 – 7x + 3 = 0
(2x – 1)(x – 3) = 0
x = ½ or x = 3
y( ½ ) = 4 ½ + 1 = 3
y( 3 ) = 4 3+ 1 = 13
Complete solution ( ½ , 3) and (3, 13)
K S Sandhu
Simultaneous equationsSolve y = x2 + 3x - 18 and y = 2x + 2
x2 + 3x - 18 = 2x + 2
x2 + x - 20 = 0
(x + 5)(x - 4) = 0
x = -5 or x = 4
y( -5 ) = 2 (-5) + 2 = -8
y( 4 ) = 2 4+ 2 = 10
Complete solution ( -5 , -8) and (4, 10)
K S Sandhu
Simultaneous equationsShow that the simultaneous equations y = x2 + 3x - 18 and y = 2x – 20 have no real solution.
x2 + 3x - 18 = 2x - 20
x2 + x + 2 = 0
a =1, b = 1 and c = 2
b2 – 4ac = 1 – 4 1 2 = - 7 < 0
The line and the curve do not intersect i.e. no real solutions
K S Sandhu
Simultaneous equationsThe graph of y = 2x + k meets the graph of y = x2 + 3x – 18 at only one point. Find the value of k.
x2 + 3x - 18 = 2x + k
x2 + x - k - 18 = 0
a =1, b = 1 and c = -k - 18
b2 – 4ac = 1 – 4 1 (- k – 18) = 0
1 + 4k + 72 = 0
k = -73/4
K S Sandhu
Simultaneous equationsIt is given that x and y satisfy the simultaneous equations:
y – x = 4 and 2x2 + xy + y2 = 8
y – x = 4 y = x + 4
2x2 + xy + y2 = 8
4x2 + 12x + 8 = 0
(i) Show that x2 + 3x + 2 = 0 (ii) Solve the equations.
2x2 + x(x + 4) + (x + 4)2 = 8
2x2 + x2 + 4x + x2 + 8x + 16 = 8
x2 + 3x + 2 = 0
(i)
(x + 2)(x + 1)= 0(ii) x = - 2 or x = - 1
Complete solution (-2, 2) or (-1, 3)
K S Sandhu
Simultaneous equationsSolve the simultaneous equations.
x + y = 1 and x2 - xy + y2 = 7
x + y = 1 y = 1 - x
x2 – x(1 – x) + (1 – x)2 = 7
3x2 - 3x - 6 = 0
x2 – x + x2 + 1 – 2x + x2 = 7
x2 - x - 2 = 0
(x + 1)(x - 2)= 0
x = - 1 or x = 2
Complete solution (-1, 2) or (2, -1)
K S Sandhu
Substitute for x from (1) into (2)
Factorise and solve for y
Substitute for y in (1)
Solve the simultaneous equationsx + y = 2 (1) x2 + 2y2 = 11 (2)
(2 – y)2 + 2y2 = 11
4 – 4y + y2 + 2y2 = 11
(3y – 7)(y + 1) = 0
3y2 – 4y – 7 = 0
73 or 1y y
7 7 13 3 3When , 2y x x
When 1, ( 1) 2 3y x x 71
3 3The solutions are , and 3, 1.x y x y
K S Sandhu
Substitute for x from (2) into (1)
Factorise and solve for y
Substitute for x in (2)
Solve the simultaneous equations y = x2 (1) x + y = 6 (2)
y = (6 – y)2
y = 36 – 12y + y2
(y – 4)(y - 9) = 0
y2 – 13y + 36 = 0
When 4, 4 6 2y x x When 9, 9 6 3y x x
The solutions are 2, 4 and 3, 9.x y x y
y = 4 or y = 9