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3 years ago

How many solutions are there to the equation below 9x + 27 = 9( x + 2 ) + 9

Mathematics

2 answers:

Nataly [62]3 years ago

7 0

Let's solve the equation:

9x+27 = 9(x+2)+9 ← Distribute 9 to the x and 2

9x+27 = 9x+18+9 ← Combine like terms

9x+27 = 9x + 27 ← Subtract 27 from both sides

9x = 9x

Infinitely many solutions would be correct because no matter what x is, it will always equal each other the both sides of the equation because it is 9 times x on both sides.

Mama L [17]3 years ago

6 0

9x + 27 = 9(x + 2) + 9

Simplify both sides of equation:

9x + 27 = 9(x + 2) + 9

9x + 27 = (9)(x) + (9)(2) + 9 (Distribute)

9x + 27 = 9x + 18 + 9

9x + 27 = (9x) + ( 18 + 9)

(Combine Like Terms:)

9x + 27 = 9x + 27

Subtract 9x from both sides:

9x + 27 - 9x = 9x + 27 - 9x

27 = 27

Subtract 27 from both sides:

27 - 27 = 27 - 27

0 = 0

Equations is always true

Answer: All real numbers are solutions: (Infinitely many solutions:)

Hope that helps!!! ( Answer: Letter Choice (B), Infinitely many solutions.).

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How does the graph of f(x)=3lx+2l+4 relate to its parent function?

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$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\

\begin{array}{rllll} 
% left side templates
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}
\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\end{array}$

$\bf \begin{array}{llll}


\bullet \textit{ vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\\\
\qquad if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}
\end{array}$

now, with that template above in mind, let's see this one

$\bf parent\implies f(x)=|x|
\\\\\\
\begin{array}{lllcclll}
f(x)=&3|&1x&+2|&+4\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}$

A=3, B=1, shrunk by AB or 3 units, about 1/3
C=2, horizontal shift by C/B or 2/1 or just 2, to the left
D=4, vertical shift upwards of 4 units

check the picture below