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
[tex]\dfrac{ -60x }{ -60 } = \dfrac{ -106 }{ -60 }[/tex]
Simplify
[tex]x = \dfrac{ 53 }{ 30 }[/tex]
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"
[tex]\mathrm{So\:it\:would\:look\:like\:this:\ y = -5 \left( \dfrac{ 53 }{ 30 } \right) +8}[/tex]
[tex]\mathrm{Now\:lets\:solve\:for\:"y"\:then}[/tex]
[tex]y = -5 \left( \dfrac{ 53 }{ 30 } \right) +8}[/tex]
[tex]\mathrm{Express\: -5 \times \dfrac{ 53 }{ 30 }\:as\:a\:single\:fraction}[/tex]
[tex]y = \dfrac{ -5 \times 53 }{ 30 } +8[/tex]
[tex]\mathrm{Multiply\:-5 \:and\:53\:to\:get\:-265 }[/tex]
[tex]y = \dfrac{ -265 }{ 30 } +8[/tex]
[tex]\mathrm{Simplify\: \dfrac{ -265 }{ 30 } \:,by\:dividing\:both\:-265\:and\:30\:by\:5} }[/tex]
[tex]y = \dfrac{ -265 \div 5 }{ 30 \div 5 } +8[/tex]
[tex]\mathrm{Simplify}[/tex]
[tex]y = - \dfrac{ 53 }{ 6 } +8[/tex]
[tex]\mathrm{Turn\:8\:into\:a\:fraction\:that\:has\:the\:same\:denominator\:as\: - \dfrac{ 53 }{ 6 }}[/tex]
[tex]\mathrm{Multiples\:of\:1: \:1,2,3,4,5,6}[/tex]
[tex]\mathrm{Multiples\:of\:6: \:6,12,18,24,30,36,42,48}[/tex]
[tex]\mathrm{Convert\:8\:to\:fraction\:\dfrac{ 48 }{ 6 }}[/tex]
[tex]y = - \dfrac{ 53 }{ 6 } + \dfrac{ 48 }{ 6 }[/tex]
[tex]\mathrm{Since\: - \dfrac{ 53 }{ 6 }\:have\:the\:same\:denominator\:,\:add\:them\:by\:adding\:their\:numerators}[/tex]
[tex]y = \dfrac{ -53+48 }{ 6 }[/tex]
[tex]\mathrm{Add\: -53 \: and\: 48\: to\: get\: -5}[/tex]
[tex]y = - \dfrac{ 5 }{ 6 }[/tex]
[tex]\mathrm{The\:solution\:is\:the\:ordered\:pair\:(\dfrac{ 53 }{ 30 }, - \dfrac{ 5 }{ 6 })}[/tex]
So there is only one solution to the equation.
