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

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Algebra 2

Quadratic equation with complex roots

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5

. Determine the exact intersection point of lines

3 y=6 x-5

and

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

Convert to standard form: Write both equations in the standard form of a line, which is $y = mx + b$ , where $m$ is the slope and $b$ is the y-intercept. $\newline$ For the first equation, $3y = 6x - 5$ , divide by $3$ to isolate $y$ : $\newline$ $y = \frac{6}{3}x - \frac{5}{3}$ $\newline$ $y = 2x - \frac{5}{3}$
Isolate variables: Write the second equation in the standard form. $\newline$ For the second equation, $3x = 7y - \left(\frac{109}{3}\right)$ , divide by $3$ to isolate $x$ : $\newline$ $x = \left(\frac{7}{3}\right)y - \left(\frac{109}{9}\right)$ $\newline$ Now, solve for $y$ by multiplying both sides by $\frac{3}{7}$ : $\newline$ $y = \left(\frac{3}{7}\right)x + \left(\frac{109}{21}\right)$
Set equations equal: Set the two equations equal to each other to find the point of intersection. $\newline$ $2x - \frac{5}{3} = \frac{3}{7}x + \frac{109}{21}$
Solve for x: Solve for x by getting all $x$ terms on one side and constants on the other. $\newline$ Multiply both sides by $21$ to clear the fractions: $\newline$ $42x - 35 = 9x + 109$
Substitute $x$ : Subtract $9x$ from both sides to get all $x$ terms on one side: $\newline$ $42x - 9x - 35 = 109$ $\newline$ $33x - 35 = 109$
Simplify $y$ : Add $35$ to both sides to isolate the $x$ term: $\newline$ $33x = 109 + 35$ $\newline$ $33x = 144$
Find intersection point: Divide both sides by $33$ to solve for $x$ : $\newline$ $x = \frac{144}{33}$ $\newline$ $x = 4.36363636...$ $\newline$ However, since we are looking for the exact value, we should keep it as a fraction: $\newline$ $x = \frac{144}{33}$
Find intersection point: Divide both sides by $33$ to solve for $x$ : $x = \frac{144}{33}$ $x = 4.36363636...$ However, since we are looking for the exact value, we should keep it as a fraction: $x = \frac{144}{33}$ Substitute the value of $x$ back into one of the original equations to solve for $y$ . We'll use the first equation $y = 2x - \frac{5}{3}$ : $y = 2\left(\frac{144}{33}\right) - \left(\frac{5}{3}\right)$
Find intersection point: Divide both sides by $33$ to solve for $x$ : $\newline$ $x = \frac{144}{33}$ $\newline$ $x = 4.36363636...$ $\newline$ However, since we are looking for the exact value, we should keep it as a fraction: $\newline$ $x = \frac{144}{33}$ Substitute the value of $x$ back into one of the original equations to solve for $y$ . We'll use the first equation $y = 2x - \frac{5}{3}$ : $\newline$ $y = 2\left(\frac{144}{33}\right) - \left(\frac{5}{3}\right)$ Simplify the expression for $y$ : $\newline$ $y = \left(\frac{288}{33}\right) - \left(\frac{5}{3}\right)$ $\newline$ To subtract these fractions, they need a common denominator. The common denominator is $33$ : $\newline$ $y = \left(\frac{288}{33}\right) - \left(\frac{55}{33}\right)$
Find intersection point: Divide both sides by $33$ to solve for $x$ : $\newline$ $x = \frac{144}{33}$ $\newline$ $x = 4.36363636...$ $\newline$ However, since we are looking for the exact value, we should keep it as a fraction: $\newline$ $x = \frac{144}{33}$ Substitute the value of $x$ back into one of the original equations to solve for $y$ . We'll use the first equation $y = 2x - \frac{5}{3}$ : $\newline$ $y = 2\left(\frac{144}{33}\right) - \left(\frac{5}{3}\right)$ Simplify the expression for $y$ : $\newline$ $y = \left(\frac{288}{33}\right) - \left(\frac{5}{3}\right)$ $\newline$ To subtract these fractions, they need a common denominator. The common denominator is $33$ : $\newline$ $y = \left(\frac{288}{33}\right) - \left(\frac{55}{33}\right)$ Subtract the fractions to solve for $y$ : $\newline$ $y = \frac{288 - 55}{33}$ $\newline$ $y = \frac{233}{33}$
Find intersection point: Divide both sides by $33$ to solve for $x$ : $\newline$ $x = \frac{144}{33}$ $\newline$ $x = 4.36363636...$ $\newline$ However, since we are looking for the exact value, we should keep it as a fraction: $\newline$ $x = \frac{144}{33}$ Substitute the value of $x$ back into one of the original equations to solve for $y$ . We'll use the first equation $y = 2x - \frac{5}{3}$ : $\newline$ $y = 2\left(\frac{144}{33}\right) - \left(\frac{5}{3}\right)$ Simplify the expression for $y$ : $\newline$ $y = \left(\frac{288}{33}\right) - \left(\frac{5}{3}\right)$ $\newline$ To subtract these fractions, they need a common denominator. The common denominator is $33$ : $\newline$ $y = \left(\frac{288}{33}\right) - \left(\frac{55}{33}\right)$ Subtract the fractions to solve for $y$ : $\newline$ $y = \frac{288 - 55}{33}$ $\newline$ $y = \frac{233}{33}$ Write the intersection point using the values of $x$ and $y$ : $\newline$ The intersection point is $x$ $1$ .