solve the equation
x+15/2 = 2x
Answers
Answer:
Solution−
Given equation is
\begin{gathered} \rm \: {x}^{2} + x - {y}^{2} - 3y - 2 = 0 \\ \end{gathered}
x
2
+x−y
2
−3y−2=0
can be rewritten as
\begin{gathered} \rm \: {x}^{2} + x - ( {y}^{2} + 3y + 2) = 0 \\ \end{gathered}
x
2
+x−(y
2
+3y+2)=0
\begin{gathered} \rm \: {x}^{2} + x - ( {y}^{2} + 2y + y+ 2) = 0 \\ \end{gathered}
x
2
+x−(y
2
+2y+y+2)=0
\begin{gathered}\rm \: {x}^{2} + x - [y(y - 2) - 1(y - 2)] = 0 \\ \end{gathered}
x
2
+x−[y(y−2)−1(y−2)]=0
\begin{gathered}\rm \: {x}^{2} + x - (y + 1)(y + 2) = 0 \\ \end{gathered}
x
2
+x−(y+1)(y+2)=0
can be further rewritten as
\begin{gathered}\rm \: {x}^{2} + \red{(2 - 1)}x - (y + 1)(y + 2) = 0 \\ \end{gathered}
x
2
+(2−1)x−(y+1)(y+2)=0
\begin{gathered}\rm \: {x}^{2} + \red{(2 - 1 + y - y)}x - (y + 1)(y + 2) = 0 \\ \end{gathered}
x
2
+(2−1+y−y)x−(y+1)(y+2)=0
\begin{gathered}\rm \: {x}^{2} + \red{(y + 2 - y - 1)}x - (y + 1)(y + 2) = 0 \\ \end{gathered}
x
2
+(y+2−y−1)x−(y+1)(y+2)=0
\begin{gathered}\rm \: {x}^{2} + \red{[(y + 2) - (y + 1)]}x - (y + 1)(y + 2) = 0 \\ \end{gathered}
x
2
+[(y+2)−(y+1)]x−(y+1)(y+2)=0
\begin{gathered}\rm \: {x}^{2} + (y + 2)x - (y + 1)x - (y + 1)(y + 2) = 0 \\ \end{gathered}
x
2
+(y+2)x−(y+1)x−(y+1)(y+2)=0
\begin{gathered}\rm \: x[x + y + 2] - (y + 1)[x + y + 2] = 0 \\ \end{gathered}
x[x+y+2]−(y+1)[x+y+2]=0
\begin{gathered}\rm \: [x + y + 2][x - y - 1] = 0 \\ \end{gathered}
[x+y+2][x−y−1]=0
\begin{gathered}\rm\implies \:x = - (y + 2) \: \: or \: \: x = y + 1 \\ \end{gathered}
⟹x=−(y+2)orx=y+1
Hence,
\begin{gathered}\begin{gathered}\begin{gathered}\bf\: \rm\implies \:Solution \: is \: \begin{cases} &\sf{x = y + 1} \\ \\ &\sf{x = - y - 2} \end{cases}\end{gathered}\end{gathered}\end{gathered}
⟹Solutionis
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⎪
⎪
⎨
⎪
⎪
⎧
x=y+1
x=−y−2