It is known that $y = ax^2 + bx^3$; when $x = 2$, $y = 5.6$, and when $x = 3$, $y = 25$. Find the values of a and b. $$5.6 = 4a + 8b$$
$$25 = 9a + 27b$$
$$5.6 - 25 = ( 4a + 8b ) - ( 9a - 27b )$$
$$-19.4 = -5a - 19b$$
I'm stuck at that point.
$$a = -1.36$$ $$b = 1.38$$
 A: METHOD 1: ELIMINATION
Multiplying the first equation by $9$ gives us that $$50.4=36a+72b$$ Multiplying the second equation by $4$ gives us that $$100=36a+108b$$ Now subtract these two equations which gives us $b$. Now put this back in the first equation to get $a$. 
METHOD 2: SUBSTITUTION
Dividing the first equation by $4$ gives us that $a+2b=1.4$. So $a=1.4-2b$. Then put this in the second equation, which gives us a equation of the form $9b-12.4=0$, which we can use to find $b$. Now put this back in $a=1.4-2b$ to get $a$. 
In Any a case, we get that $$a=-\frac{61}{45}$$ $$b=\frac{62}{45}$$
Though my answer differs from yours, it can be demonstrated that your answer is incorrect: if we put $a=-1.36, b=1.38$, then we get $25.02$ in the second equation, not $25$. 
A: It is easier to just change one of the equations to be in terms of $a$ or $b$, so a "substitution" approach rather than "elimination".
So from your first equation: 
$4 = 5.6 - 8b$
$a = 1.4 - 2b$
Substitute this into $25 = 9a + 27b$ to get
$25 = 9(1.4-2b) + 27b$
Solve this to get $b$.
Then put this back into $a = 1.4 -2b$
