Question

$$\left. \begin{array} { l } { 2 x ^ { 2 } + 6 x - 10 = 3 } \\ { 3 x ^ { 2 } + 4 = 0 , 1 } \end{array} \right.$$

Answer

$$2*e*s*h^3*o*w*t^3*a*x^2+12*e^2*IM*q*u*a^2*l*f*r*(m*a*x-1)^2$$

Solution


Regroup terms.
\[2shhhowttta{x}^{2}e+{(max-1)}^{2}\times 3equal\imath f\times 4are\]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[2s{h}^{1+1+1}ow{t}^{1+1+1}a{x}^{2}e+{(max-1)}^{2}\times 3equal\imath f\times 4are\]
Simplify  \(1+1\)  to  \(2\).
\[2s{h}^{2+1}ow{t}^{2+1}a{x}^{2}e+{(max-1)}^{2}\times 3equal\imath f\times 4are\]
Simplify  \(2+1\)  to  \(3\).
\[2s{h}^{3}ow{t}^{3}a{x}^{2}e+{(max-1)}^{2}\times 3equal\imath f\times 4are\]
Regroup terms.
\[2es{h}^{3}ow{t}^{3}a{x}^{2}+{(max-1)}^{2}\times 3equal\imath f\times 4are\]
Take out the constants.
\[2es{h}^{3}ow{t}^{3}a{x}^{2}+(3\times 4)quaalfr{(max-1)}^{2}e\imath e\]
Simplify  \(3\times 4\)  to  \(12\).
\[2es{h}^{3}ow{t}^{3}a{x}^{2}+12quaalfr{(max-1)}^{2}e\imath e\]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[2es{h}^{3}ow{t}^{3}a{x}^{2}+12qu{a}^{2}lfr{(max-1)}^{2}{e}^{2}\imath \]
Regroup terms.
\[2es{h}^{3}ow{t}^{3}a{x}^{2}+12{e}^{2}\imath qu{a}^{2}lfr{(max-1)}^{2}\]