The equations arranged in order from least to greatest based on their solution is Equation B < Equation C < Equation A
For Equation A
[tex]5(x-6) + 3x = \frac{3}{4}(2x-8)[/tex]
First, clear the fraction by multiplying through by 4
[tex]4\times 5(x-6) + 4\times3x = 4\times\frac{3}{4}(2x-8) \\20(x-6) + 12x = 3(2x-8)[/tex]
Now, clear the brackets
[tex]20x-120 + 12x = 6x-24[/tex]
Collect like terms
[tex]20x+ 12x-6x = -24+120[/tex]
Then,
[tex]26x = 96[/tex]
Divide both sides by 26
[tex]\frac{26x}{26}= \frac{96}{26} \\x = 3.69[/tex]
∴ For Equation A, [tex]x =3.69[/tex]
For Equation B
[tex]2.7(5.1x+4.9) =3.2x+28.9[/tex]
First, clear the bracket
[tex]13.77x+13.23 =3.2x+28.9[/tex]
Now, collect like terms
[tex]13.77x-3.2x =28.9-13.23\\10.57x = 15.67[/tex]
Divide both sides by 10.57
[tex]\frac{10.57x}{10.57}= \frac{15.67}{10.57} \\x = 1.48[/tex]
∴ For Equation B, [tex]x=1.48[/tex]
For Equation C
[tex]5(11x-18)=3(2x+7)[/tex]
First, clear the brackets
[tex]55x-90=6x+21[/tex]
Now, collect like terms
[tex]55x-6x=90+21[/tex]
Then,
[tex]49x = 111[/tex]
Divide both sides by 49
[tex]\frac{49x}{49} = \frac{111}{49}\\x = 2.27[/tex]
∴ For Equation C, [tex]x=2.27[/tex]
For Equation A, [tex]x =3.69[/tex]
For Equation B, [tex]x=1.48[/tex]
For Equation C, [tex]x=2.27[/tex]
Arranging the solutions from the least to greatest, we get
1.48 < 2.27 < 3.69
∴ Equation B < Equation C < Equation A
Hence, the equations arranged in order from least to greatest based on their solution is Equation B < Equation C < Equation A
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