Antwoord:
# x = 3,5 en y = 1 #
OF
#x = 2,5 en y = 3 #
Uitleg:
# 2x + y = 8 ………………………………… (1) #
# 4x ^ 2 + 3y ^ 2 = 52 ………………….. (2) #
(1) # => y = 8-2x #
(2) # => 4x ^ 2 + 3 (8-2x) ^ 2 = 52 #
# => 4x ^ 2 +3 (64 - 32x + 4x ^ 2) = 52 #
# => 4x ^ 2 + 192 - 96x + 12x ^ 2 = 52 #
# => 16x ^ 2 -96x + 140 = 0 #
# => 4 (4x ^ 2 - 24x +35) = 0 #
# => 4x ^ 2 -24x +35 = 0 #
Door deze kwadratische vergelijking op te lossen, krijgen we:
# => (x-3.5) (x-2.5) = 0 #
# => x = 3.5 of x = 2.5 #
Vervang deze waarde van #X# in vergelijking (1):
Case 1: Taking # x = 3.5 #
# => 2x + y = 8 #
# => 2 (3.5) + y = 8 #
# => y = 8-7 = 1 #
OF
Case 2: Taking # x = 2.5 #
# 2 (2.5) + y = 8 #
# => y = 8- 5 = 3 #
#therefore x = 3.5 en y = 1 #
OF
#x = 2,5 en y = 3 #
Antwoord:
# y = 3 en x = 5/2 of y = 1 en x = 7/2 #
Uitleg:
# 2x + y = 8 #
# 4x ^ 2 + 3y ^ 2 = 52 #
# 2x = 8-y => 4x ^ 2 = (8-y) ^ 2 = kleur (blauw) (64-16y + y ^ 2) #
#color (blauw) (64-16y + y ^ 2) ^ 3y + 2 = 52 #
# 4j ^ 2-16y + 12 = 0 #
alles delen door 4 (voor berekeningen):
# Y ^ 2-4Y + 3 = 0 #
# Y = (4 + -sqrt (4 ^ 2-4 * 1 * 3)) / 2 #
# Y = (4 + -sqrt (4)) / 2 #
# Y = (4 ± 2) / 2 #
# y = 3 en x = 5/2 of x = 1 en x = 7/2 #
