wglenda26
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# The sum of two angles is 71∘. Angle 2 is 49∘ smaller than 2 times angle 1. How do I do this type of problem? How do I figure it out?

Let's call $\begin{Bmatrix}Angle~1&=&x\\Angle~2&=&y\end{matrix}$ OK?! so $\begin{Bmatrix}x+y&=&71\º\\y&=&2x-49\º\end{matrix}$ we can replace at the first row $\begin{Bmatrix}x+2x-49&=&71\º\\y&=&2x-49\º\end{matrix}$ $\begin{Bmatrix}3x&=&71\º+49\º\\y&=&2x-49\º\end{matrix}$ $\begin{Bmatrix}3x&=&120\º\\y&=&2x-49\º\end{matrix}$ $\begin{Bmatrix}x&=&40\º\\y&=&2x-49\º\end{matrix}$ now we can find the second angle... what we just have to do is replace the value that we find to x at the second row $\begin{Bmatrix}x&=&40\º\\y&=&2*40\º-49\º\end{matrix}$ $\begin{Bmatrix}x&=&40\º\\y&=&80\º-49\º\end{matrix}$ $\begin{Bmatrix}x&=&40\º\\y&=&31\º\end{matrix}$ $\boxed{\boxed{\therefore\begin{Bmatrix}Angle~1&=&40\º\\Angle~2&=&31\º\end{matrix}}}$