1. Solve for \( x \) if: \( \begin{array}{ll}1.1 .1 & (2 x-1)(x+5)=0 \\ 1.1 .2 & x(3 x+13)=11 \\ 1.1 .3 & 1+\sqrt{2 x+1}=x \\ 1.1 .4 & 2^{x \sqrt{x}}=2^{27} \\ 1.1 .5 & x^{2}<-2 x+15 \\ \text { Solve for } x & \text { and } y \text { simultaneously if: } \\ 3-y+2 x=0 \text { and } 6 x+4 y^{2}=3+5 x y\end{array} \)
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The Deep Dive
To tackle this math puzzle, let's dive into each equation one by one! For the equation \( (2x-1)(x+5)=0 \), we find that either \( 2x-1=0 \) or \( x+5=0 \). Solving these gives us \( x=\frac{1}{2} \) and \( x=-5 \). Next, the equation \( x(3x+13)=11 \) transforms into a standard quadratic equation when rearranged: \( 3x^2 + 13x - 11 = 0 \). Using the quadratic formula, we can find two values for \( x \). Moving on to \( 1+\sqrt{2x+1}=x \), we isolate \( \sqrt{2x+1} \) and square both sides to simplify and solve for \( x \). For \( 2^{x\sqrt{x}}=2^{27} \), since the bases are equal, we set the exponents equal to each other: \( x\sqrt{x}=27 \) or \( x^{3/2}=27 \), allowing us to find \( x \). Now for the inequality \( x^2<-2x+15 \), we rearrange it into a quadratic form and analyze the roots for potential valid intervals of \( x \). And finally, for the simultaneous equations \( 3-y+2x=0 \) and \( 6x+4y^2=3+5xy \), we can express \( y \) in terms of \( x \) from the first equation, substitute into the second equation, and solve it step by step! Start solving one by one, and enjoy the ride through the mathematical wilderness! Keep your pencil sharp!