{:[2(x-(1)/(3))-(3)/(2)(y-(1)/(6))=0],[3(y-(1)/(2))+(8)/(3)(x-(1)/(6))=0]:} Consider the system of equations. If (x,y) is the solution to the system, what is the value of the sum of x and y ? Choose 1 answer: (A) (5)/(6) (B) (25)/(36) (C) (2)/(3) (D) None of the above

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{:[2(x-(1)/(3))-(3)/(2)(y-(1)/(6))=0],[3(y-(1)/(2))+(8)/(3)(x-(1)/(6))=0]:} Consider the system of equations. If  (x,y) is the solution to the system, what is the value of the sum of  x and  y ? Choose 1 answer: (A)  (5)/(6) (B)  (25)/(36) (C)  (2)/(3) (D) None of the above

Bytelearn · Accepted Answer

Write Equations: First, let's write down the system of equations: $$ \begin{cases}  2\left(x-\frac{1}{3}\right)-\frac{3}{2}\left(y-\frac{1}{6}\right)=0 \ 3\left(y-\frac{1}{2}\right)+\frac{8}{3}\left(x-\frac{1}{6}\right)=0 \end{cases} $$Simplify Equations: Next, we simplify each equation by distributing the constants and moving all terms to one side: For the first equation: $$ 2x - \frac{2}{3} - \frac{3}{2}y + \frac{1}{4} = 0 $$ $$ 2x - \frac{3}{2}y - \frac{2}{3} + \frac{1}{4} = 0 $$ $$ 2x - \frac{3}{2}y - \frac{8}{12} + \frac{3}{12} = 0 $$ $$ 2x - \frac{3}{2}y - \frac{5}{12} = 0 $$ $$ 2x - \frac{3}{2}y = \frac{5}{12} $$ For the second equation: $$ 3y - \frac{3}{2} + \frac{8}{3}x - \frac{8}{18} = 0 $$ $$ 3y + \frac{8}{3}x - \frac{3}{2} - \frac{4}{9} = 0 $$ $$ 3y + \frac{8}{3}x - \frac{27}{18} - \frac{4}{9} = 0 $$ $$ 3y + \frac{8}{3}x - \frac{27}{18} - \frac{8}{18} = 0 $$ $$ 3y + \frac{8}{3}x - \frac{35}{18} = 0 $$ $$ 3y + \frac{8}{3}x = \frac{35}{18} $$Standard Form: Now we have the system of equations in a more standard form: $$ \begin{cases}  2x - \frac{3}{2}y = \frac{5}{12} \ 3y + \frac{8}{3}x = \frac{35}{18} \end{cases} $$Elimination Method: To solve the system, we can use either substitution or elimination. Let's use the elimination method. We'll multiply the first equation by 3 and the second equation by 2 to eliminate y: $$ \begin{cases}  (2x - \frac{3}{2}y) \cdot 3 = \frac{5}{12} \cdot 3 \ (3y + \frac{8}{3}x) \cdot 2 = \frac{35}{18} \cdot 2 \end{cases} $$ This gives us: $$ \begin{cases}  6x - \frac{9}{2}y = \frac{15}{12} \ 6y + \frac{16}{3}x = \frac{35}{9} \end{cases} $$Simplify Equations: Now we simplify the equations: $$ \begin{cases}  6x - \frac{9}{2}y = \frac{5}{4} \ 6y + \frac{16}{3}x = \frac{70}{18} \end{cases} $$ $$ \begin{cases}  6x - \frac{9}{2}y = \frac{5}{4} \ 6y + \frac{16}{3}x = \frac{35}{9} \end{cases} $$Common Denominator: Next, we'll multiply the second equation by 2 to get the same denominator for x in both equations: $$ \begin{cases}  6x - \frac{9}{2}y = \frac{5}{4} \ (6y + \frac{16}{3}x) \cdot 2 = (\frac{35}{9}) \cdot 2 \end{cases} $$ This gives us: $$ \begin{cases}  6x - \frac{9}{2}y = \frac{5}{4} \ 12y + \frac{32}{3}x = \frac{70}{9} \end{cases} $$Combine Equations: Now we have the system: $$ \begin{cases}  6x - \frac{9}{2}y = \frac{5}{4} \ 12y + \frac{32}{3}x = \frac{70}{9} \end{cases} $$ We can multiply the first equation by 2 to match the denominators: $$ \begin{cases}  (6x - \frac{9}{2}y) \cdot 2 = (\frac{5}{4}) \cdot 2 \ 12y + \frac{32}{3}x = \frac{70}{9} \end{cases} $$ This gives us: $$ \begin{cases}  12x - 9y = \frac{5}{2} \ 12y + \frac{32}{3}x = \frac{70}{9} \end{cases} $$Solve for x: Now we can add the two equations together to eliminate y: $$ (12x - 9y) + (12y + \frac{32}{3}x) = \frac{5}{2} + \frac{70}{9} $$ $$ 12x + \frac{32}{3}x = \frac{5}{2} + \frac{70}{9} + 9y - 12y $$ $$ 12x + \frac{32}{3}x = \frac{5}{2} + \frac{70}{9} $$Substitute x into Equation: To combine the x terms, we need a common denominator. We multiply 12x by 3/3: $$ \frac{36}{3}x + \frac{32}{3}x = \frac{5}{2} + \frac{70}{9} $$ $$ \frac{68}{3}x = \frac{5}{2} + \frac{70}{9} $$Solve for y: Now we find a common denominator for the right side of the equation: $$ \frac{68}{3}x = \frac{5 \cdot 9}{2 \cdot 9} + \frac{70 \cdot 2}{9 \cdot 2} $$ $$ \frac{68}{3}x = \frac{45}{18} + \frac{140}{18} $$ $$ \frac{68}{3}x = \frac{185}{18} $$Calculate Sum: Now we solve for x: $$ x = \frac{185}{18} \cdot \frac{3}{68} $$ $$ x = \frac{185}{68} \cdot \frac{1}{6} $$ $$ x = \frac{185}{408} $$ $$ x = \frac{5}{12} $$Now that we have x, we can substitute it back into one of the original equations to find y. Let's use the first equation: $$ 2x - \frac{3}{2}y = \frac{5}{12} $$ $$ 2 \cdot \frac{5}{12} - \frac{3}{2}y = \frac{5}{12} $$ $$ \frac{5}{6} - \frac{3}{2}y = \frac{5}{12} $$Now we solve for y: $$ -\frac{3}{2}y = \frac{5}{12} - \frac{5}{6} $$ $$ -\frac{3}{2}y = \frac{5}{12} - \frac{10}{12} $$ $$ -\frac{3}{2}y = -\frac{5}{12} $$ $$ y = -\frac{5}{12} \cdot -\frac{2}{3} $$ $$ y = \frac{5}{12} \cdot \frac{2}{3} $$ $$ y = \frac{5}{18} $$Now we have both x and y: $$ x = \frac{5}{12} $$ $$ y = \frac{5}{18} $$ The sum of x and y is: $$ x + y = \frac{5}{12} + \frac{5}{18} $$To add the fractions, we find a common denominator: $$ x + y = \frac{5 \cdot 3}{12 \cdot 3} + \frac{5 \cdot 2}{18 \cdot 2} $$ $$ x + y = \frac{15}{36} + \frac{10}{36} $$ $$ x + y = \frac{25}{36} $$

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