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Delvig [45]
2 years ago
10

How many solutions does the system of equations below have?

Mathematics
1 answer:
soldier1979 [14.2K]2 years ago
5 0

Answer:

One solution                    

Step-by-step explanation:

5x + y = 8

15x + 15y = 14

Lets solve using substitution, first we need to turn "5x + = 8" into "y = mx + b" or slope - intercept form

So we solve for "y" in the equation "5x + y = 8"

5x + y = 8

Step 1: Subtract 5x from both sides.

5x + y − 5x = 8 − 5x

Step 2: 5x subtracted by 5x cancel out and "8 - 5x" are flipped

y = −5x + 8

Now we can solve using substitution:

We substitute "-5x + 8" into the equation "15x + 15y = 14" for y

So it would look like this:

15x + 15(-5x + 8) = 14

Now we just solve for x

15x + (15)(−5x) + (15)(8) = 14(Distribute)

15x − 75x + 120 = 14

(15x − 75x) + (120) = 14(Combine Like Terms)

−60x + 120 = 14

Step 2: Subtract 120 from both sides.

−60x + 120 − 120 = 14 − 120

−60x = −106

Divide both sides by -60

\dfrac{ -60x  }{ -60  }   =   \dfrac{ -106  }{ -60  }

Simplify

x =   \dfrac{ 53  }{ 30  }

Now that we know the value of x, we can solve for y in any of the equations, but let's use the equation "y = −5x + 8"

\mathrm{So\:it\:would\:look\:like\:this:\ y =  -5 \left(  \dfrac{ 53  }{ 30  }    \right)  +8}

\mathrm{Now\:lets\:solve\:for\:"y"\:then}

y =  -5 \left(  \dfrac{ 53  }{ 30  }    \right)  +8}

\mathrm{Express\: -5 \times   \dfrac{ 53  }{ 30  }\:as\:a\:single\:fraction}

y =   \dfrac{ -5 \times  53  }{ 30  }  +8

\mathrm{Multiply\:-5 \:and\:53\:to\:get\:-265 }

y =   \dfrac{ -265  }{ 30  }  +8

\mathrm{Simplify\:  \dfrac{ -265  }{ 30  }    \:,by\:dividing\:both\:-265\:and\:30\:by\:5} }

y =   \dfrac{ -265 \div  5  }{ 30 \div  5  }  +8

\mathrm{Simplify}

y =  - \dfrac{ 53  }{ 6  }  +8

\mathrm{Turn\:8\:into\:a\:fraction\:that\:has\:the\:same\:denominator\:as\: - \dfrac{ 53  }{ 6  }}

\mathrm{Multiples\:of\:1: \:1,2,3,4,5,6}

\mathrm{Multiples\:of\:6: \:6,12,18,24,30,36,42,48}

\mathrm{Convert\:8\:to\:fraction\:\dfrac{ 48  }{ 6  }}

y =  - \dfrac{ 53  }{ 6  }  + \dfrac{ 48  }{ 6  }

\mathrm{Since\: - \dfrac{ 53  }{ 6  }\:have\:the\:same\:denominator\:,\:add\:them\:by\:adding\:their\:numerators}

y =   \dfrac{ -53+48  }{ 6  }

\mathrm{Add\: -53 \: and\: 48\: to\: get\:  -5}

y =  - \dfrac{ 5  }{ 6  }

\mathrm{The\:solution\:is\:the\:ordered\:pair\:(\dfrac{ 53  }{ 30  }, - \dfrac{ 5  }{ 6  })}

So there is only one solution to the equation.

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