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NikAS [45]
2 years ago
15

Can you guys help? Need to find x, BA, DA

Mathematics
1 answer:
7nadin3 [17]2 years ago
4 0

Answer:

x=2,BA=DA which is7

Step-by-step explanation:

5x-3=3x+1

both sides share an angle and B=D right angles

5x-3x=1+3

2x=4

x=4÷2

x=2

BA=5x-3

BA=5(2)-3

BA=10-3

BA=7

DA=3x+1

DA=3(2)+1

DA=6+1

DA=7

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What is the solution of ?<img src="https://tex.z-dn.net/?f=%5Cfrac%7B5%7D%7B2%7D%20x-7%3D%5Cfrac%7B3%7D%7B4%7D%20x%2B14" id="Tex
Nonamiya [84]

Combine the terms by multiplying into a single fraction.

\bf{\dfrac{5}{2}x-7=\dfrac{3}{4}x+14   }

Find the common denominator.

\bf{\dfrac{5x}{2}-7=\dfrac{3}{4}x+14   }

Combine fractions with the lowest common denominator.

\bf{\dfrac{5x(-7)}{2}=\dfrac{3}{4}x+14   }

Multiply the numbers.

\bf{\dfrac{5x-14}{2}=\dfrac{3}{4}x+14   }

Combine the multiplied terms into a single fraction

\bf{\dfrac{5x-14}{2}=\dfrac{3x}{4}+14   }

Find the common denominator.

\bf{\dfrac{5x-14}{2}=\dfrac{3x}{4}+\dfrac{4\cdot14}{4}    }

Combine fractions with the lowest common denominator.

\bf{\dfrac{5x-14}{2}=\dfrac{3x+4\cdot14}{4}   }

Multiply the numbers.

\bf{\dfrac{5x-14}{2}=\dfrac{3x+56}{4}   }

Eliminate the denominators of the fractions.

\bf{4\cdot\dfrac{5x-14}{2}=4\cdot\dfrac{3x+56}{4}   }

Cancel the multiplied terms that are in the denominator.

\bf{2(5x-14)=3x+56}

To distribute.

\bf{10x-28=3x+56 }

Add 28 to both sides.

\bf{10x-28+28=3x+56+28 }

Simplify

\bf{10x=3x+84 }

Subtract 3x from both sides.

\bf{10x-3x=3x+84-3x }

Simplify

\bf{7x=84 }

Divide both sides by the same factor.

\bf{x=\dfrac{84}{7} }

Simplify

\bf{x=12 \ \ \ === > \ \ \ Answer}

↓

<h3>Verification</h3>

Let x=12.

  1. \bf{\dfrac{5}{2}\times12-7=\dfrac{3}{4}\times12+14   }
  2. \bf{\dfrac{5\times12}{2}-7=\dfrac{3}{4}\times12+14   }
  3. \bf{\dfrac{60}{2}-7=\dfrac{3}{4}\times12+14   }
  4. \bf{30-7=\dfrac{3}{4}\times12+14   }
  5. \bf{30-7=\dfrac{3\times12}{4}+14   }
  6. \bf{30-7=\dfrac{36}{4}+14   }
  7. \bf{30-7=9+14   }
  8. \bf{23=9+14 }
  9. \bf{23=23}

Checked ✅

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Here's how I got this answer:
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