Solution 2.2:2c

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We can simplify the left-hand side in the equation by expanding the squares using the squaring rule:
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We can simplify the left-hand side in the equation by expanding the squares using the squaring rule
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<math>\begin{align}
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& \left( x+3 \right)^{2}-\left( x-5 \right)^{2}=\left( x^{2}+2\centerdot 3x+3^{2} \right)-\left( x^{2}-2\centerdot 5x+5^{2} \right) \\
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& =x^{2}+6x+9-x^{2}+10x-25=16x-16 \\
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\end{align}</math>
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{{Displayed math||<math>\begin{align}
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(x+3)^{2}-(x-5)^{2}
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&= (x^{2}+2\cdot 3x+3^{2})-(x^{2}-2\cdot 5x+5^{2})\\[5pt]
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&= x^{2}+6x+9-x^{2}+10x-25\\[5pt]
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&=16x-16\,\textrm{.}
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\end{align}</math>}}
Thus, the equation is
Thus, the equation is
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{{Displayed math||<math>16x-16=6x+4\,\textrm{.}</math>}}
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<math>16x-16=6x+4</math>
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Now, move all ''x'''s to the left-hand side (subtract 6''x'' from both sides) and the constants to the right-hand side (add 16 to both sides)
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Now, move all "
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<math>x</math>
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"s to the left-hand side (subtract
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<math>6x</math>
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from both sides) and the constants to the right-hand side (add
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<math>16</math>
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to both sides)
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<math>\begin{align}
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& 16x-6x=4+16 \\
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& 10x=20 \\
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\end{align}</math>
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Divide both sides by
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<math>10</math>
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to get the answer
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<math>x=\frac{20}{10}=2</math>
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{{Displayed math||<math>\begin{align}
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16x-6x&=4+16\,,\\[5pt]
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10x&=20\,\textrm{.}
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\end{align}</math>}}
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Finally, we check that
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Divide both sides by 10 to get the answer
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<math>x=2</math>
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satisfies the equation in the exercise:
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LHS =
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{{Displayed math||<math>x=\frac{20}{10}=2\,\textrm{.}</math>}}
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<math>\left( 2+3 \right)^{2}-\left( 2-5 \right)^{2}=5^{2}-\left( -3 \right)^{2}=25-9=16</math>
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Finally, we check that <math>x=2</math> satisfies the equation in the exercise
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RHS =
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{{Displayed math||<math>\begin{align}
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<math>6\centerdot 2+4=12+4=16</math>
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\text{LHS} &= (2+3)^{2}-(2-5)^{2} = 5^{2}-(-3)^{2} = 25-9 = 16,\\[5pt]
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\text{RHS} &= 6\cdot 2+4 = 12+4 = 16\,\textrm{.}
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\end{align}</math>}}

Current revision

We can simplify the left-hand side in the equation by expanding the squares using the squaring rule

\displaystyle \begin{align}

(x+3)^{2}-(x-5)^{2} &= (x^{2}+2\cdot 3x+3^{2})-(x^{2}-2\cdot 5x+5^{2})\\[5pt] &= x^{2}+6x+9-x^{2}+10x-25\\[5pt] &=16x-16\,\textrm{.} \end{align}

Thus, the equation is

\displaystyle 16x-16=6x+4\,\textrm{.}

Now, move all x's to the left-hand side (subtract 6x from both sides) and the constants to the right-hand side (add 16 to both sides)

\displaystyle \begin{align}

16x-6x&=4+16\,,\\[5pt] 10x&=20\,\textrm{.} \end{align}

Divide both sides by 10 to get the answer

\displaystyle x=\frac{20}{10}=2\,\textrm{.}

Finally, we check that \displaystyle x=2 satisfies the equation in the exercise

\displaystyle \begin{align}

\text{LHS} &= (2+3)^{2}-(2-5)^{2} = 5^{2}-(-3)^{2} = 25-9 = 16,\\[5pt] \text{RHS} &= 6\cdot 2+4 = 12+4 = 16\,\textrm{.} \end{align}

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