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allochka39001 [22]
2 years ago
12

How do you solve 32 - 6x = 53?

Mathematics
2 answers:
katrin2010 [14]2 years ago
7 0

\large\boxed{x=-\frac{7}{2}}

To solve for x, we need to isolate it on one side of the equation.

The most important part of this is knowing that whatever we do to one side of the equation, we must also do to the other.

Subtract 32 from both sides of the equation.

\begin{aligned}32-32-6x&=53-32\\-6x&=21\end{aligned}

Divide both sides of the equation by -6.

\begin{aligned}\frac{-6x}{-6}&=\frac{21}{-6}\\x&=\boxed{-\frac{7}{2}}\end{aligned}

Cerrena [4.2K]2 years ago
3 0

Answer:

x = - 3.5

Step-by-step explanation:

32 - 6x = 53 ( subtract 32 from both sides )

- 6x = 21 ( divide both sides by - 6 )

x = \frac{21}{-6} = - \frac{7}{2} = - 3.5

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The length of a rectangle is 5 metres less than twice the breadth. If the perimeter is 50 meters,find the length and breadth
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<h3><u>S</u><u> </u><u>O</u><u> </u><u>L</u><u> </u><u>U</u><u> </u><u>T</u><u> </u><u>I</u><u> </u><u>O</u><u> </u><u>N</u><u> </u><u>:</u></h3>

As per the given question, it is stated that the length of a rectangle is 5 m less than twice the breadth.

Assumption : Let us assume the length as "l" and width as "b". So,

\twoheadrightarrow \quad\sf{ Length =2(Width)-5}

\twoheadrightarrow \quad\sf{ \ell=(2b-5) \; m}

Also, we are given that the perimeter of the rectangle is 50 m. Basically, we need to apply here the formula of perimeter of rectangle which will act as a linear equation here.

\\ \twoheadrightarrow \quad\sf{ Perimeter_{(Rectangle)} = 2(\ell +b) } \\

  • <em>l</em> denotes length
  • <em>b</em> denotes breadth

\\ \twoheadrightarrow \quad\sf{50= 2(2b-5+b)} \\

\\ \twoheadrightarrow \quad\sf{50= 2(3b-5)} \\

\\ \twoheadrightarrow \quad\sf{50= 6b - 10} \\

\\ \twoheadrightarrow \quad\sf{50+10= 6b} \\

\\ \twoheadrightarrow \quad\sf{60= 6b} \\

\\ \twoheadrightarrow \quad\sf{\cancel{\dfrac{60}{6}}=b} \\

\\ \twoheadrightarrow \quad\underline{\bf{10\; m = Width }} \\

Now, finding the length. According to the question,

\twoheadrightarrow \quad\sf{ \ell=(2b-5) \; m}

\twoheadrightarrow \quad\sf{ \ell=2(10)-5\; m}

\twoheadrightarrow \quad\sf{ \ell=20-5\; m}

\\ \twoheadrightarrow \quad\underline{\bf{15\; m = Length }} \\

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