Determine the exact intersection point of lines $3y=6x-5$ and $3x=7y-\frac{109}{3}$

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Q. Determine the exact intersection point of lines

3y=6x-5

and

3x=7y-\frac{109}{3}

Set Up Equations: To find the intersection point of two lines, we need to solve the system of equations given by the two lines. The first equation is $3y = 6x - 5$ , and the second equation is $3x = 7y - \frac{109}{3}.$
Simplify Equations: Let's simplify the equations to make them easier to work with. For the first equation, we can divide both sides by $3$ to get $y = 2x - \frac{5}{3}$ . For the second equation, we can also divide both sides by $3$ to get $x = \left(\frac{7}{3}\right)y - \frac{109}{9}$ .
Substitute and Solve for y: Now we have two equations in slope-intercept form: $\newline$ $1$ ) $y = 2x - \frac{5}{3}$ $\newline$ $2$ ) $x = \left(\frac{7}{3}\right)y - \frac{109}{9}$ $\newline$ We can substitute the expression for $x$ from the second equation into the first equation to solve for $y$ .
Combine Like Terms for $y$ : Substituting $x$ from the second equation into the first equation gives us: $\newline$ $y = 2\left(\frac{7}{3}y - \frac{109}{9}\right) - \frac{5}{3}$ $\newline$ Now we need to distribute the $2$ and simplify the equation to solve for $y$ .
Combine Terms on Right Side: Distributing the $2$ gives us: $\newline$ $y = \left(\frac{14}{3}\right)y - \frac{218}{9} - \frac{5}{3}$ $\newline$ Now we combine like terms and solve for $y$ .
Solve for y: Combining like terms, we get: $\newline$ $y - \frac{14}{3}y = -\frac{218}{9} - \frac{5}{3}$ $\newline$ To combine the terms on the left, we need a common denominator, which is $3$ .
Substitute and Solve for x: Converting $y$ to have the same denominator as $(\frac{14}{3})y$ , we get: $\newline$ $(\frac{3}{3})y - (\frac{14}{3})y = \frac{-218}{9} - \frac{5}{3}$ $\newline$ Now we can subtract the fractions on the left side.
Perform Multiplication and Subtraction: Subtracting the fractions on the left side gives us: $\newline$ $-\frac{11}{3}y = -\frac{218}{9} - \frac{5}{3}$ $\newline$ Now we need to combine the terms on the right side.
Subtract Fractions: Combining the terms on the right side, we need a common denominator, which is $9$ . $\newline$ $-\frac{5}{3}$ is the same as $-\frac{15}{9}$ , so we have: $\newline$ $-\left(\frac{11}{3}\right)y = -\frac{218}{9} - \frac{15}{9}$ $\newline$ Now we can add the fractions on the right side.
Subtract Numerators: Adding the fractions on the right side gives us: $\newline$ $-\frac{11}{3}y = -\frac{233}{9}$ $\newline$ Now we can solve for $y$ by dividing both sides by $-\frac{11}{3}$ .
Find $x$ : Dividing both sides by $-\left(\frac{11}{3}\right)$ gives us: $\newline$ $y = \frac{-\frac{233}{9}}{-\left(\frac{11}{3}\right)}$ $\newline$ To divide by a fraction, we multiply by its reciprocal.
Intersection Point: Multiplying by the reciprocal of $-(11/3)$ , we get: $\newline$ y = $(-233/9) \times (-3/11)$ $\newline$ Now we can multiply the numerators and denominators.
Intersection Point: Multiplying by the reciprocal of $-(11/3)$ , we get: $\newline$ y = $(-233/9) \times (-3/11)$ $\newline$ Now we can multiply the numerators and denominators.Multiplying the numerators and denominators gives us: $\newline$ y = $(233 \times 3) / (9 \times 11)$ $\newline$ Now we can simplify the fraction.
Intersection Point: Multiplying by the reciprocal of $-\frac{11}{3}$ , we get: $\newline$ y = $-\frac{233}{9}$ * $-\frac{3}{11}$ $\newline$ Now we can multiply the numerators and denominators.Multiplying the numerators and denominators gives us: $\newline$ y = $\frac{233 \times 3}{9 \times 11}$ $\newline$ Now we can simplify the fraction.Simplifying the fraction, we get: $\newline$ y = \frac{ $699$ }{ $99$ } $\newline$ Both the numerator and denominator are divisible by $99$ .
Intersection Point: Multiplying by the reciprocal of $-(11/3)$ , we get: $\newline$ y = $(-233/9) \times (-3/11)$ $\newline$ Now we can multiply the numerators and denominators.Multiplying the numerators and denominators gives us: $\newline$ y = $(233 \times 3) / (9 \times 11)$ $\newline$ Now we can simplify the fraction.Simplifying the fraction, we get: $\newline$ y = $699 / 99$ $\newline$ Both the numerator and denominator are divisible by $99$ .Dividing both the numerator and denominator by $99$ , we get: $\newline$ y = $7$ $\newline$ Now that we have the value of y, we can substitute it back into one of the original equations to find x.
Intersection Point: Multiplying by the reciprocal of $-(11/3)$ , we get: $\newline$ y = $(-233/9) \times (-3/11)$ $\newline$ Now we can multiply the numerators and denominators.Multiplying the numerators and denominators gives us: $\newline$ y = $(233 \times 3) / (9 \times 11)$ $\newline$ Now we can simplify the fraction.Simplifying the fraction, we get: $\newline$ y = $699 / 99$ $\newline$ Both the numerator and denominator are divisible by $99$ .Dividing both the numerator and denominator by $99$ , we get: $\newline$ y = $7$ $\newline$ Now that we have the value of y, we can substitute it back into one of the original equations to find x.Substituting $y = 7$ into the second equation $x = (7/3)y - 109/9$ , we get: $\newline$ x = $(7/3)\times7 - 109/9$ $\newline$ Now we need to perform the multiplication and subtraction to find x.
Intersection Point: Multiplying by the reciprocal of $-\frac{11}{3}$ , we get: $\newline$ y = $-\frac{233}{9}$ * $-\frac{3}{11}$ $\newline$ Now we can multiply the numerators and denominators.Multiplying the numerators and denominators gives us: $\newline$ y = $\frac{233 \times 3}{9 \times 11}$ $\newline$ Now we can simplify the fraction.Simplifying the fraction, we get: $\newline$ y = \frac{ $699$ }{ $99$ } $\newline$ Both the numerator and denominator are divisible by $99$ .Dividing both the numerator and denominator by $99$ , we get: $\newline$ y = $7$ $\newline$ Now that we have the value of y, we can substitute it back into one of the original equations to find x.Substituting y = $7$ into the second equation x = $\frac{7}{3}$ y - \frac{ $109$ }{ $9$ }\, we get: $\newline$ x = $\frac{7}{3}$ * $7$ - \frac{ $109$ }{ $9$ } $\newline$ Now we need to perform the multiplication and subtraction to find x.Performing the multiplication and subtraction, we get: $\newline$ x = \frac{ $49$ }{ $3$ } - \frac{ $109$ }{ $9$ } $\newline$ To subtract these fractions, we need a common denominator, which is $9$ .
Intersection Point: Multiplying by the reciprocal of $-\frac{11}{3}$ , we get: $\newline$ y = $-\frac{233}{9}$ * $-\frac{3}{11}$ $\newline$ Now we can multiply the numerators and denominators.Multiplying the numerators and denominators gives us: $\newline$ y = $\frac{233 \times 3}{9 \times 11}$ $\newline$ Now we can simplify the fraction.Simplifying the fraction, we get: $\newline$ y = \frac{ $699$ }{ $99$ } $\newline$ Both the numerator and denominator are divisible by $99$ .Dividing both the numerator and denominator by $99$ , we get: $\newline$ y = $7$ $\newline$ Now that we have the value of y, we can substitute it back into one of the original equations to find x.Substituting $y = 7$ into the second equation $x = \frac{7}{3}y - \frac{109}{9}$ , we get: $\newline$ x = $\frac{7}{3}$ \times $7$ - \frac{ $109$ }{ $9$ } $\newline$ Now we need to perform the multiplication and subtraction to find x.Performing the multiplication and subtraction, we get: $\newline$ x = \frac{ $49$ }{ $3$ } - \frac{ $109$ }{ $9$ } $\newline$ To subtract these fractions, we need a common denominator, which is $-\frac{233}{9}$ $0$ .Converting $-\frac{233}{9}$ $1$ to have the same denominator as $-\frac{233}{9}$ $2$ , we get: $\newline$ x = \frac{ $147$ }{ $9$ } - \frac{ $109$ }{ $9$ } $\newline$ Now we can subtract the fractions.
Intersection Point: Multiplying by the reciprocal of $-\frac{11}{3}$ , we get: $\newline$ y = $-\frac{233}{9}$ * $-\frac{3}{11}$ $\newline$ Now we can multiply the numerators and denominators.Multiplying the numerators and denominators gives us: $\newline$ y = $\frac{233 \times 3}{9 \times 11}$ $\newline$ Now we can simplify the fraction.Simplifying the fraction, we get: $\newline$ y = \frac{ $699$ }{ $99$ } $\newline$ Both the numerator and denominator are divisible by $99$ .Dividing both the numerator and denominator by $99$ , we get: $\newline$ y = $7$ $\newline$ Now that we have the value of y, we can substitute it back into one of the original equations to find x.Substituting $y = 7$ into the second equation $x = \frac{7}{3}y - \frac{109}{9}$ , we get: $\newline$ x = $\frac{7}{3}$ $7$ - \frac{ $109$ }{ $9$ } $\newline$ Now we need to perform the multiplication and subtraction to find x.Performing the multiplication and subtraction, we get: $\newline$ x = \frac{ $49$ }{ $3$ } - \frac{ $109$ }{ $9$ } $\newline$ To subtract these fractions, we need a common denominator, which is $-\frac{233}{9}$ $1$ .Converting $-\frac{233}{9}$ $2$ to have the same denominator as $-\frac{233}{9}$ $3$ , we get: $\newline$ x = \frac{ $147$ }{ $9$ } - \frac{ $109$ }{ $9$ } $\newline$ Now we can subtract the fractions.Subtracting the fractions gives us: $\newline$ x = \frac{( $147$ - $109$ )}{ $9$ } $\newline$ Now we can subtract the numerators.
Intersection Point: Multiplying by the reciprocal of $-\frac{11}{3}$ , we get: $\newline$ $y = \left(-\frac{233}{9}\right) \times \left(-\frac{3}{11}\right)$ $\newline$ Now we can multiply the numerators and denominators.Multiplying the numerators and denominators gives us: $\newline$ $y = \frac{233 \times 3}{9 \times 11}$ $\newline$ Now we can simplify the fraction.Simplifying the fraction, we get: $\newline$ $y = \frac{699}{99}$ $\newline$ Both the numerator and denominator are divisible by $99$ .Dividing both the numerator and denominator by $99$ , we get: $\newline$ $y = 7$ $\newline$ Now that we have the value of $y$ , we can substitute it back into one of the original equations to find $x$ .Substituting $y = 7$ into the second equation $x = \frac{7}{3}y - \frac{109}{9}$ , we get: $\newline$ $x = \left(\frac{7}{3}\right)\times7 - \frac{109}{9}$ $\newline$ Now we need to perform the multiplication and subtraction to find $x$ .Performing the multiplication and subtraction, we get: $\newline$ $x = \frac{49}{3} - \frac{109}{9}$ $\newline$ To subtract these fractions, we need a common denominator, which is $9$ .Converting $\frac{49}{3}$ to have the same denominator as $99$ $0$ , we get: $\newline$ $x = \frac{147}{9} - \frac{109}{9}$ $\newline$ Now we can subtract the fractions.Subtracting the fractions gives us: $\newline$ $x = \frac{147 - 109}{9}$ $\newline$ Now we can subtract the numerators.Subtracting the numerators gives us: $\newline$ $x = \frac{38}{9}$ $\newline$ Now we have the value of $x$ .
Intersection Point: Multiplying by the reciprocal of $-\frac{11}{3}$ , we get: $\newline$ $y = \left(-\frac{233}{9}\right) \times \left(-\frac{3}{11}\right)$ $\newline$ Now we can multiply the numerators and denominators.Multiplying the numerators and denominators gives us: $\newline$ $y = \frac{233 \times 3}{9 \times 11}$ $\newline$ Now we can simplify the fraction.Simplifying the fraction, we get: $\newline$ $y = \frac{699}{99}$ $\newline$ Both the numerator and denominator are divisible by $99$ .Dividing both the numerator and denominator by $99$ , we get: $\newline$ $y = 7$ $\newline$ Now that we have the value of $y$ , we can substitute it back into one of the original equations to find $x$ .Substituting $y = 7$ into the second equation $x = \frac{7}{3}y - \frac{109}{9}$ , we get: $\newline$ $x = \left(\frac{7}{3}\right)\times7 - \frac{109}{9}$ $\newline$ Now we need to perform the multiplication and subtraction to find $x$ .Performing the multiplication and subtraction, we get: $\newline$ $x = \frac{49}{3} - \frac{109}{9}$ $\newline$ To subtract these fractions, we need a common denominator, which is $9$ .Converting $\frac{49}{3}$ to have the same denominator as $99$ $0$ , we get: $\newline$ $x = \frac{147}{9} - \frac{109}{9}$ $\newline$ Now we can subtract the fractions.Subtracting the fractions gives us: $\newline$ $x = \frac{147 - 109}{9}$ $\newline$ Now we can subtract the numerators.Subtracting the numerators gives us: $\newline$ $x = \frac{38}{9}$ $\newline$ Now we have the value of $x$ .The intersection point of the two lines is the point $99$ $2$ , which we have found to be $99$ $3$ .

Determine the exact intersection point of lines 3y=6x−53y=6x-53y=6x−5 and 3x=7y−10933x=7y-\frac{109}{3}3x=7y−3109​

Determine the exact intersection point of lines $3y=6x-5$ and $3x=7y-\frac{109}{3}$