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A level Math question. Help

love_lavlavender

2x+3y=5, x^2+3xy=4

Reply 1

mqb2766

Original post by love_lavlavender

2x+3y=5, x^2+3xy=4

Just do a substitution and solve the quadratic?

Reply 2

MohummudQAMAR

Original post by love_lavlavender

2x+3y=5, x^2+3xy=4

make first equation equal to 3y
and theres a 3xy or (3y)(x) in the other equation.
just always remember to always try to work with only one unknown variable in an equation

Reply 3

jahanzaib9

Original post by love_lavlavender

2x+3y=5, x^2+3xy=4

just a normal pair of simultaneous equations.

best way to go about this is to make y the subject in equation 1 and substituting that in equation 2. we wouldnt make x the subject and sub it in as we have to square it which would make it longer.

you'd get y = (5 - 2x)/3 for equation 1
next just sub that into equation 2:
x² + 3x((5-2x)/3) = 4
notice that the 3's will cancel, after simplifying you get:
x² + x(5-2x) = 4
x² + 5x - 2x² - 4 = 0
-x² + 5x - 4 = 0
this can also be written as x² -5x + 4 = 0
now you solve this quadratic:
you get x = 4 and x = -1
now we need to find a corresponding solution for each value of x

using the equation: 2x + 3y = 5, y = (5-2x)/3
when x is 4
y = (5-2(4))/3
y = -1

and when x is 1
y = (5-2(1))/3
y = 1

so your solutions are:

x=4,y=-1
and
x=1,y=1

Reply 4

love_lavlavender