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3 years ago

How many solutions are there to the equation below 6x+15=6(x+3)

Mathematics

2 answers:

Shalnov [3]3 years ago

8 0

Answer:

The equation has Zero solution.

Step-by-step explanation:

Consider the provided equation.

$6x+15=6(x+3)$

Apply the distributive property: $a(b+c)=ab+ac$

$6x+15=6x+18$

Now isolate the variables by subtracting 6 from both the sides.

$6x+15-6x=6x+18-6x$

$15=18$

Which is not true.

Hence, the equation has no solution or the equation has Zero solution.

cupoosta [38]3 years ago

7 0

6x + 15 = 6 (x + 3)

6x + 15 = 6x + 18

6x = 6x + 3

0 = 3

So, there are no solutions to this equation.

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Answer:

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4 years ago

Read 2 more answers

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blagie [28]

A.
$5x-4=-2(3x+2) \ \ \ \ \ \ \ \ \ |\hbox{expand the bracket} \\
5x-4=-2 \times 3x-2 \times 2 \\
5x-4=-6x-4 \ \ \ \ \ \ \ \ \ \ \ \ \ |\hbox{add 6x to both sides} \\
11x-4=-4 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\hbox{add 4 to both sides} \\
11x=0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\hbox{divide both sides by 11} \\
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B.
$3(2x-4)=5x-1 \\
6x-12=5x-1 \\
\boxed{11x-12=-1} \Leftarrow \hbox{the first mistake} \\
11x=11 \\
\boxed{x=11} \Leftarrow \hbox{the second mistake}$

Megan's solution isn't correct.
The first mistake: she subtracted 5x from the right-hand side of the equation, but added 5x to the left-hand side.
The second mistake: she divided the right-hand side of the equation by 11, but didn't divide the left-hand side.

The correct solution:
$3(2x-4)=5x-1 \ \ \ \ \ \ \ \ \ \ \ |\hbox{expand the bracket} \\
3 \times 2x+3 \times (-4)=5x-1 \\ 6x-12=5x-1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\hbox{subtract 5x from both sides} \\
x-12=-1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\hbox{add 12 to both sides} \\
x=11$

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posledela

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WARRIOR [948]

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