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Solve for all values of x. 2x+(7)/(x+6)=3x Answer Attemptiout of 2 Additional Solution (C) No Solution x=square

Pergunta

Solve for all values of x.
2x+(7)/(x+6)=3x
Answer Attemptiout of 2
Additional Solution (C) No Solution
x=square

Solve for all values of x. 2x+(7)/(x+6)=3x Answer Attemptiout of 2 Additional Solution (C) No Solution x=square

Solução

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GustavoProfissional · Tutor por 6 anos

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To solve the equation \(2x + \frac{7}{x+6} = 3x\), we need to isolate \(x\). Here are the steps:<br /><br />1. Subtract \(2x\) from both sides of the equation:<br /> \[<br /> \frac{7}{x+6} = 3x - 2x<br /> \]<br /> Simplifying the right side, we get:<br /> \[<br /> \frac{7}{x+6} = x<br /> \]<br /><br />2. Multiply both sides by \(x + 6\) to eliminate the fraction:<br /> \[<br /> 7 = x(x + 6)<br /> \]<br /><br />3. Expand the right side:<br /> \[<br /> 7 = x^2 + 6x<br /> \]<br /><br />4. Rearrange the equation to standard quadratic form:<br /> \[<br /> x^2 + 6x - 7 = 0<br /> \]<br /><br />5. Solve the quadratic equation using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 6\), and \(c = -7\):<br /> \[<br /> x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot (-7)}}{2 \cdot 1}<br /> \]<br /> \[<br /> x = \frac{-6 \pm \sqrt{36 + 28}}{2}<br /> \]<br /> \[<br /> x = \frac{-6 \pm \sqrt{64}}{2}<br /> \]<br /> \[<br /> x = \frac{-6 \pm 8}{2}<br /> \]<br /><br />6. This gives us two solutions:<br /> \[<br /> x = \frac{-6 + 8}{2} = \frac{2}{2} = 1<br /> \]<br /> \[<br /> x = \frac{-6 - 8}{2} = \frac{-14}{2} = -7<br /> \]<br /><br />So, the solutions to the equation are:<br />\[<br />x = 1 \quad \text{or} \quad x = -7<br />\]<br /><br />We should verify these solutions in the original equation to ensure they are correct.<br /><br />For \(x = 1\):<br />\[<br />2(1) + \frac{7}{1+6} = 3(1)<br />\]<br />\[<br />2 + \frac{7}{7} = 3<br />\]<br />\[<br />2 + 1 = 3<br />\]<br />This is true.<br /><br />For \(x = -7\):<br />\[<br />2(-7) + \frac{7}{-7+6} = 3(-7)<br />\]<br />\[<br />-14 + \frac{7}{-1} = -21<br />\]<br />\[<br />-14 - 7 = -21<br />\]<br />This is also true.<br /><br />Therefore, the solutions are:<br />\[<br />x = 1 \quad \text{or} \quad x = -7<br />\]
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