First, combine like terms on the left-hand side of the equation.

$\displaystyle 4x-x=3x$

Now, the equation is

$\displaystyle 3x+5=x+3$

From here, subtract $\displaystyle x$ from both sides.

  $\displaystyle 3x+5=x+3$

$\displaystyle -x$           $\displaystyle -x$

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$\displaystyle 2x+5=3$

Next, subtract five to both sides.

$\displaystyle 2x+5=3$

       $\displaystyle -5$   $\displaystyle -5$

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$\displaystyle 2x=-2$

Finally, divide both sides of the equation by two.

$\displaystyle \frac{2x}{2}=\frac{-2}{2}$

$\displaystyle x=-1$

To solve for $\displaystyle x$, first combine like terms by adding $\displaystyle 2x$ to both sides.

$\displaystyle 7x-5=51$

Next, add $\displaystyle 5$ to both sides.

$\displaystyle 7x=56$

From here, divide by $\displaystyle 7$ to solve for $\displaystyle x$.

$\displaystyle x=8$

To solve for $\displaystyle x$, square both terms to cancel out the square root sign on the left-hand side.

$\displaystyle (\sqrt{x-2})^2=(1)^2$

$\displaystyle x-2=1$

Next, add two to both sides of the equation.

$\displaystyle x-2=1$

     $\displaystyle +2$  $\displaystyle +2$

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$\displaystyle x=3$

From here, check for extraneous solutions by substituting the value found for $\displaystyle x$ into the original equation.

$\displaystyle \sqrt{x-2}=1$

$\displaystyle \\ \sqrt{3-2}=1 \\\sqrt{1}=1 \\\pm1=1$

Since the square root of one is either positive or negative one, the solution found is verified.

To solve for $\displaystyle x$, first subtract one from both sides.

    $\displaystyle 1-\sqrt{x+1}=5$

$\displaystyle -1$                        $\displaystyle -1$

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$\displaystyle -\sqrt{x+1}=4$

From here, divide both sides by negative one.

$\displaystyle -\frac{\sqrt{x+1}}{-1}=\frac{4}{-1}$

$\displaystyle \sqrt{x+1}=-4$

Next, square both sides to cancel the square root sign on the left-hand side.

$\displaystyle (\sqrt{x+1})^2=(-4)^2$

$\displaystyle x+1=16$

Now, subtract one from both sides to solve for $\displaystyle x$.

$\displaystyle x+1=16$

     $\displaystyle -1$     $\displaystyle -1$

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$\displaystyle x=15$

Lastly, check for extraneous solutions by substituting the value found for $\displaystyle x$ into the original equation.

$\displaystyle 1-\sqrt{x+1}=5$

$\displaystyle \\1-\sqrt{15+1}=5 \\1-\sqrt{16}=5 \\1-\pm4=5$

Thus, the answer is verified.

To solve for $\displaystyle x$, add two to both sides of the equation.

$\displaystyle \frac{1}{x}-2=10$

     $\displaystyle +2$     $\displaystyle +2$

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$\displaystyle \frac{1}{x}=12$

Now, multiply by $\displaystyle x$ on both sides. This will move the variable from the denominator on one side to the numerator of the other side.

$\displaystyle x\cdot \frac{1}{x}=12\cdot x$

$\displaystyle 1=12x$

Lastly, divide both sides by twelve.

$\displaystyle \frac{1}{12}=\frac{12x}{12}$

The twelve in the numerator and the twelve in the denominator cancel out thus, solving for $\displaystyle x$.

$\displaystyle \frac{1}{12}=x$

To solve for $\displaystyle x$, first subtract four from both sides so all constants are on one side.

$\displaystyle \sqrt{x}+4=3$

         $\displaystyle -4$  $\displaystyle -4$

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$\displaystyle \sqrt{x}=-1$

Now, square both sides to cancel the square root sign on the left-hand side.

$\displaystyle (\sqrt{x})^2=(-1)^2$

$\displaystyle x=1$

Recall that when a negative number is squared the result is always a positive value.

From here, check for extraneous solutions by substituting in the value found for $\displaystyle x$.

$\displaystyle \\\sqrt{x}+4=3 \\\sqrt{1}+4=3 \\\pm1+4=3 \\-1+4=3$

Since the square root of a value results in a positive and a negative value, our solution is verified.

To solve for $\displaystyle x$ start by squaring both sides to eliminate the square root sign.

$\displaystyle \left( \sqrt{\frac{x}{2}}\right)^2=(6)^2$

$\displaystyle \frac{x}{2}=36$

From here multiply both sides by two. On the left-hand side, the two in the numerator will cancel out the two in the denominator.

$\displaystyle 2\cdot \frac{x}{2}=36\cdot 2$

$\displaystyle x=72$

From here, check for extraneous solutions by substituting in the value found for $\displaystyle x$ into the original equation.

$\displaystyle \sqrt{\frac{x}{2}}=6$

$\displaystyle \sqrt{\frac{72}{2}}=6$

$\displaystyle \sqrt{36}=6$

$\displaystyle \pm6=6$

Thus, the solution is verified.

To solve for $\displaystyle x$, first add two to both sides.

$\displaystyle \frac{3}{x}-2=5$

     $\displaystyle +2$   $\displaystyle +2$

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$\displaystyle \frac{3}{x}=7$

From here, multiply both sides by $\displaystyle x$. 

$\displaystyle x\cdot \frac{3}{x}=7\cdot x$

$\displaystyle 3=7x$

Now divide by seven to solve for $\displaystyle x$.

$\displaystyle \frac{3}{7}=\frac{7x}{7}$

The seven in the numerator and seven in the denominator cancel out, thus solving for $\displaystyle x$.

$\displaystyle \frac{3}{7}=x$

To solve for $\displaystyle x$, first square both sides of the equation. Squaring a square root sign will cancel them out.

$\displaystyle (\sqrt{5-x})^2=(2)^2$

$\displaystyle 5-x=4$

Now, subtract five from both sides to get all constants on one side of the equation while all variables are on the other side of the equation.

   $\displaystyle 5-x=4$

$\displaystyle -5$           $\displaystyle -5$

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$\displaystyle -x=-1$

From here, divide by negative one on both sides.

$\displaystyle \frac{-x}{-1}=\frac{-1}{-1}$

$\displaystyle x=1$

Lastly, check for extraneous solutions by substituting in the value found for $\displaystyle x$ into the original equation.

$\displaystyle \\\sqrt{5-1}=2 \\\sqrt{4}=2 \\\pm2=2$

Thus, the solution is verified.

To solve for $\displaystyle x$, simply multiply both sides of the equation by two.

$\displaystyle \frac{x}{2}=8$

$\displaystyle 2\cdot \frac{x}{2}=8\cdot 2$

The two in the numerator cancels the two in the denominator, thus solving for $\displaystyle x$.

$\displaystyle x=16$