Consider the system of equations. If $(x, y)$ is the solution to the system, what is the value of the sum of $x$ and $y$ ? $\newline$ $2(x-\frac{1}{3})-\frac{3}{2}(y-\frac{1}{6})=0$ $\newline$ $3(y-\frac{1}{2})+\frac{8}{3}(x-\frac{1}{6})=0$

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Find the magnitude of a vector scalar multiple

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Q. 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

\newline

2(x-\frac{1}{3})-\frac{3}{2}(y-\frac{1}{6})=0

\newline

3(y-\frac{1}{2})+\frac{8}{3}(x-\frac{1}{6})=0

Write Equations: Write down the system of equations. $\newline$ The system of equations is: $\newline$ $2(x - \frac{1}{3}) - \frac{3}{2}(y - \frac{1}{6}) = 0$ $\newline$ $3(y - \frac{1}{2}) + \frac{8}{3}(x - \frac{1}{6}) = 0$
Simplify Equations: Simplify each equation by distributing the multiplication and combining like terms. $\newline$ For the first equation: $\newline$ $2(x - \frac{1}{3}) - \frac{3}{2}(y - \frac{1}{6}) = 2x - \frac{2}{3} - \frac{3}{2}y + \frac{1}{4}$ $\newline$ Combine like terms: $\newline$ $2x - \frac{3}{2}y - \frac{2}{3} + \frac{1}{4} = 0$ $\newline$ Convert $-\frac{2}{3}$ and $\frac{1}{4}$ to have a common denominator of $12$ : $\newline$ $2x - \frac{3}{2}y - \frac{8}{12} + \frac{3}{12} = 0$ $\newline$ $2x - \frac{3}{2}y - \frac{5}{12} = 0$ $\newline$ Multiply through by $12$ to clear the fraction: $\newline$ $24x - 18y - 5 = 0$
Elimination Method: Simplify the second equation in the same way. $\newline$ For the second equation: $\newline$ $3(y - \frac{1}{2}) + \frac{8}{3}(x - \frac{1}{6}) = 3y - \frac{3}{2} + \frac{8}{3}x - \frac{8}{18}$ $\newline$ Combine like terms and convert fractions to have a common denominator: $\newline$ $3y - \frac{3}{2} + \frac{8}{3}x - \frac{4}{9} = 0$ $\newline$ Multiply through by $18$ to clear the fractions: $\newline$ $54y - 27 + 48x - 8 = 0$ $\newline$ $48x + 54y - 35 = 0$
Subtract Equations: Now we have a system of two linear equations with no fractions: $\newline$ $24x - 18y - 5 = 0$ $\newline$ $48x + 54y - 35 = 0$ $\newline$ We can use the method of substitution or elimination to solve this system. Let's use the elimination method by multiplying the first equation by $2$ to align the coefficients of $x$ . $\newline$ First equation multiplied by $2$ : $\newline$ $48x - 36y - 10 = 0$
Solve for y: Subtract the modified first equation from the second equation to eliminate $x$ .
$(48x + 54y - 35) - (48x - 36y - 10) = 0$
$48x + 54y - 35 - 48x + 36y + 10 = 0$
Combine like terms:
$90y - 25 = 0$
Substitute for $x$ : Solve for $y$ . $\newline$ Add $25$ to both sides: $\newline$ $90y = 25$ $\newline$ Divide by $90$ : $\newline$ $y = \frac{25}{90}$ $\newline$ Simplify the fraction: $\newline$ $y = \frac{5}{18}$
Find Sum: Substitute $y = \frac{5}{18}$ into one of the original equations to solve for $x$ . Let's use the first original equation: $\newline$ $24x - 18\left(\frac{5}{18}\right) - 5 = 0$ $\newline$ Simplify the multiplication: $\newline$ $24x - 5 - 5 = 0$ $\newline$ Combine like terms: $\newline$ $24x - 10 = 0$ $\newline$ Add $10$ to both sides: $\newline$ $24x = 10$ $\newline$ Divide by $24$ : $\newline$ $x = \frac{10}{24}$ $\newline$ Simplify the fraction: $\newline$ $x = \frac{5}{12}$
Find Sum: Substitute $y = \frac{5}{18}$ into one of the original equations to solve for $x$ . Let's use the first original equation: $\newline$ $24x - 18\left(\frac{5}{18}\right) - 5 = 0$ $\newline$ Simplify the multiplication: $\newline$ $24x - 5 - 5 = 0$ $\newline$ Combine like terms: $\newline$ $24x - 10 = 0$ $\newline$ Add $10$ to both sides: $\newline$ $24x = 10$ $\newline$ Divide by $24$ : $\newline$ $x = \frac{10}{24}$ $\newline$ Simplify the fraction: $\newline$ $x = \frac{5}{12}$ Find the sum of $x$ and $y$ . $\newline$ $x + y = \frac{5}{12} + \frac{5}{18}$ $\newline$ To add these fractions, find a common denominator, which is $36$ : $\newline$ $x + y = \frac{15}{36} + \frac{10}{36}$ $\newline$ Combine the fractions: $\newline$ $x + y = \frac{25}{36}$