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Zanzabum
1 year ago
12

Megan buys 3 bracelets and 3 necklaces. Each bracelet costs 5. Megan pays on 40 and gets 4 change. what is the cost of one neckl

ace?
Mathematics
1 answer:
GREYUIT [131]1 year ago
6 0
The necklaces cost $3 each
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HELP ME PLEASE!!
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\bold{\huge{\underline{ Solution }}}

<h3><u>Given </u><u>:</u><u>-</u></h3>

  • Here, We have given two equations that is ,
  • \sf{ x + 2y = 13\:\: and\:\: x + 5y = 28}

<h3><u>To </u><u>Find </u><u>:</u><u>-</u></h3>

  • We have to find the value of x and y.

<h3><u>Concept</u><u> </u><u>used </u><u>:</u><u>-</u></h3>

  • Linear equations are those equations which having highest power of degree as 1 .
  • Linear equations can be solved by subsitute method, elimination method and cross multiplication method.
  • Here, I have used substitution method to make calculation easier.
  • In subsitute method, you have to subsitute the value of one variable of eq(1) in eq(2) .

<h3><u>Let's </u><u>Begin </u><u>:</u><u>-</u></h3>

<u>Here</u><u>,</u><u> </u>

  • We have two equations, let consider these two equations as eq(1) and eq(2) that is,

\sf{ x + 2y = 13 ...eq(1)}

\sf{ x + 5y = 28 ...eq(2)}

<u>By </u><u>solving </u><u>eq(</u><u>1</u><u>)</u><u> </u><u>:</u><u>-</u>

\sf{ x + 2y = 13 }

\sf{ x  = 13 - 2y ...eq(3)}

<u>Subsitute </u><u>eq(</u><u>3</u><u>)</u><u> </u><u>in </u><u>eq(</u><u>2</u><u>)</u><u> </u><u>:</u><u>-</u>

\sf{ (13 - 2y) + 5y = 28 }

\sf{ 13 - 2y + 5y = 28 }

\sf{ 13 + 3y = 28 }

\sf{ 3y = 28 - 13 }

\sf{ 3y = 28 - 13 }

\sf{ 3y = 15 }

\sf{ y = }{\sf{\dfrac{ 15}{3}}}

\sf{ y = }{\sf{\cancel{\dfrac{ 15}{3}}}}

\bold{ y = 5 }

Thus, The value of y is 5

<h3><u>Now</u><u>, </u></h3>

<u>Subsitute </u><u>the </u><u>value </u><u>of </u><u>y </u><u>in </u><u>eq(</u><u>1</u><u>)</u><u> </u><u>:</u><u>-</u>

\sf{ x + 2(5) = 13 }

\sf{ x  + 10 = 13 }

\sf{ x   = 13 - 10 }

\bold{ x  = 3  }

Hence, The value of x and y are 5 and 3 .

\bold{\huge{\underline{ Let's \: Verify }}}

<h3><u>Equation </u><u>1</u><u> </u><u>:</u><u>-</u></h3>

\sf{ x + 2y = 13 }

\sf{ 3 + 2(5) = 13 }

\sf{ 3 + 10 = 13 }

\sf{ 13 = 13 }

\bold{ LHS = RHS }

<h3><u>Equation </u><u>2</u><u> </u><u>:</u><u>-</u><u> </u></h3>

\sf{ x + 5y = 28 }

\sf{ 3 + 5(5) = 28 }

\sf{ 3 + 25 = 28 }

\sf{ 28 = 28  }

\bold{ LHS = RHS }

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