+2440 Solving for a variable can be extremely daunting--especially in a multivariable equation such as \(5x^2-x=8y^2+\frac{9xy}{4}\). I have a few suggestions that may make this easier to do.
1. Move everything to One Side of the Equation.
This is a relatively simple step.
\(5x^2-x=8y^2+\frac{9xy}{4}\Rightarrow8y^2+\frac{9x}{4}y-5x^2+x=0\)
2. Eliminate All Instances of Fractions or Decimals
Fractions can be pesky, and there is no reason to make a hard situation worse. In this case, we can multiply both sides of the equation by 4 to eliminate the fractions. In a situation like this one, this is also relatively easy to do.
\(8y^2+\frac{9x}{4}y-5x^2+x=0\Rightarrow32y^2+9xy-20x^2+4x=0\)
3. Use a Formula to Finish it Off
This is written in the form of a quadratic, so the quadratic formula is the way to go.
| \(a=32; b=9x;c=-20x^2+4x\\ y_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\) | The only thing left to do is plug in the numbers. |
| \(y_{1,2}=\frac{-9x\pm\sqrt{(9x)^2-4*32(-20x^2+4x)}}{2*32}\) | It is time to simplify. |
| \(y_{1,2}=\frac{-9x\pm\sqrt{81x^2-128(-20x^2+4x)}}{64}\) | |
| \(y_{1,2}=\frac{-9x\pm\sqrt{81x^2+2560x^2-512x}}{64}\) | |
| \(y_{1,2}=\frac{-9x\pm\sqrt{2641x^2-512x}}{64}\) | I have now successfully solve for y. |
+2440 Both of these questions have been recently asked before, so I will redirect you to pages where others have already answered these questions because I see no reason to duplicate answers.
Question #1: https://web2.0calc.com/questions/if-abcde-x-4-edcba-what-is-the-number-abcde-as-usual
Question #2: https://web2.0calc.com/questions/steven-bakes-brownies
Cphill gets credit for the second question.
+2440 DanielCai, obviously your answer of 10.5 cannot be right since \(a<0\). I would evaluate \(g(g(g(10.5)))\) one step at a time.
| \(g(x) = \left\{ \begin{array}{cl} -x & x\le 0, \\ 2x-41 & x>0. \end{array} \right.\) | This is the function's definition. Now, let's evaluate \(g(10.5)\) |
| \(g(10.5)=2\cdot 10.5-41\\ \hspace{13mm}=21-41\\ \hspace{13mm}=-20\) | Since \(10.5>0\), evaluate the bottom function. |
| \(g(-20)=-(-20)\\ \hspace{13mm}=20\) | Since \(-20\leq0\), evaluate the top function. |
| \(g(20)=2\cdot 20-41\\ \hspace{10mm}=40-41\\ \hspace{10mm}=-1\) | Since \(20>0\), evaluate the bottom function. |
Now, let's do the same process with \(g(g(g(a)))\).
| \(g(x) = \left\{ \begin{array}{cl} -x & x\le 0, \\ 2x-41 & x>0. \end{array} \right.\) | This is the function definition. | ||
| \(g(a)=-a\) | Since \(a\leq0\) was a set parameter given at the beginning of the problem, | ||
| \(g(-a)=-2a-41\) | Since multiplying a by -1 would make \(-a>0\), so evaluate it as the bottom fraction. | ||
| I do not know if \(-2a-41>0\), so I have to consider both cases. | ||
Set these two cases equal to -1 and check both solutions.
| \(2a+41=-1\) | \(-4a-123=-1\) |
| \(2a=-42\) | \(-4a=122\) |
| \(a=-21\) | \(a=-\frac{122}{4}=-\frac{61}{2}\) |
When checking both solutions, only \(a=-\frac{61}{2}\) works.
