Question

3x^2+4y=7
5x^2+2y=12
solution of x and y?

Answer 1

lets try that again shall we
\[3x^2+4y=7\]
\[5x^2+2y=12\] double the second one get
\[10x^2+4y=24\] and then since the y's are the same set
\[10x^2-24=3x^2-7\] and solve for x

Answer 2

\[7x^2=17\]
\[x^2=\frac{17}{7}\]
\[x=\pm\sqrt{\frac{17}{7}}\]

Answer 3

3x^2+4y=7
10x^2+4y=24

7x^2=17

x=\[\pm \sqrt{17/7}\]

15x^2+20y=35
15x^2+6y=36

14y=-1

y=-1/14