Question #b8a92

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Question #b8a92

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Answer 1 · 2017-06-22T18:08:25

$see explanation$ $"see explanation"$

Explanation:

$there is no solution to the equation$ $"there is no solution to the equation"$

$x = - 9 is an extraneous solution$ $x=-9" is an extraneous solution"$

$that is a solution obtained by the simplification of the$ $"that is a solution obtained by the simplification of the "$
$process that does not satisfy the original equation$ $"process that does not satisfy the original equation"$

$As a check$ $color(blue)"As a check"$

$left side = 3 | - 18 + 7 | = 3 | - 11 | = 33$ $"left side "=3|-18+7|=3|-11|=33$

$right side = (3 \times - 9) - 6 = - 33$ $"right side " =(3xx-9)-6=-33$

$since left side \neq right side$ $"since left side " !=" right side"$

$\Rightarrow x = - 9 is an extraneous solution$ $rArrx=-9" is an extraneous solution"$

$this is a valuable lesson in the need to check solutions$ $"this is a valuable lesson in the need to check solutions"$

Answer 2 · 2017-06-22T18:11:16

Because the absolute value changes the sign and positive numbers do not equal negative numbers.

Explanation:

One way to see how the $- 9$ $-9$ is not a solution is to plug into both sides of the original

$3 | 2 \times - 9 + 7 | = 3 \times - 9 - 6$ $3abs(2xx-9+7)=3xx-9-6$

$3 | - 18 + 7 | = - 27 - 6$ $3abs(-18+7)=-27-6$

Before we go any further, you should now see these two sides cannot be equal. The left hand side will result in a positive number, but the right hand side will result in a negative number!

$3 | - 11 | = - 33$ $3abs(-11)=-33$

Remember, absolute value makes the number inside positive.

$3 \times 11 = - 33$ $3xx11=-33$

$33 \neq - 33$ $33!=-33$

If you try to solve this algebraically, you get

$3 | 2 x + 7 | = 3 x - 6$ $3abs(2x+7)=3x-6$

Divide by $3$ $3$

$| 2 x + 7 | = x - 6$ $abs(2x+7)=x-6$

The trick to solving this any more depends on the sign (positive or negative) of what's inside the absolute value brackets.

If $2 x + 7$ $2x+7$ is a negative number, then $2 x + 7 < 0$ $2x+7 < 0$ which means that $2 x < - 7$ $2x < -7$ , which means that $x < - \frac{7}{2}$ $x < -7/2$ . So for any value of $x$ $x$ less than $- \frac{7}{2}$ $-7/2$ , the inside will become POSITIVE on the left. But plugging a number less then $- \frac{7}{2}$ $-7/2$ into the right side at this point will only give NEGATIVE numbers. So the expression is false for the infinite number of values less then $- \frac{7}{2}$ $-7/2$ .

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Question #b8a92

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