Question 45·Hard·Systems of Two Linear Equations in Two Variables

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\begin{aligned} \frac{x+y}{2}+\frac{x-2y}{3} &= \frac{13}{2} \\ r(x+y)+2x-5y &= 20 \end{aligned}

In the system of equations above, $r$ is a constant. The solution to the system has an $x$ -coordinate that is $3$ less than twice its $y$ -coordinate. What is the value of $r$ ?

Your answer

When a system includes a parameter and a condition about its solution, first use any equation that does not contain the parameter together with the given condition to find the coordinates. Then substitute those coordinates into the equation containing the parameter. This avoids trying to solve a parameter-dependent system all at once.

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Write an equation for the coordinate relationship

Translate “ $3$ less than twice its $y$ -coordinate” into an equation that expresses $x$ in terms of $y$ .

Start with the equation without the parameter

Substitute your expression for $x$ into the first equation. This equation has no $r$ , so it can be used to find the coordinates first.

Substitute the coordinates

After finding $x$ and $y$ , substitute both values into the second equation and isolate $r$ .

Desmos Guide

Use algebra first

Algebra is quicker because the stated relationship between the coordinates reduces the first equation to one variable. Desmos can be used to verify the coordinate values and calculate the parameter.

Graph the two relationships for the coordinates

Multiply the first equation by $6$ and rearrange it as $y=5x-39$ . Enter y=5x-39 and y=(x+3)/2, where the second equation represents $x=2y-3$ . Click their intersection to verify the values of $x$ and $y$ .

Calculate the parameter

In a new expression line, enter (20-2(9)+5(6))/(9+6). This rearranges the second equation to isolate $r$ using the intersection coordinates.

Step-by-step Explanation

Translate the coordinate relationship

The $x$ -coordinate is $3$ less than twice the $y$ -coordinate, so

x=2y-3

Substitute this expression for $x$ into the first equation:

\begin{aligned} \frac{(2y-3)+y}{2}+\frac{(2y-3)-2y}{3} &= \frac{13}{2} \\ \frac{3y-3}{2}-1 &= \frac{13}{2} \\ 3y-5 &= 13 \end{aligned}

Thus, $y=6$ , so $x=2(6)-3=9$ .

Use the second equation to find the constant

Substitute $x=9$ and $y=6$ into the second equation:

\begin{aligned} r(9+6)+2(9)-5(6) &= 20 \\ 15r-12 &= 20 \end{aligned}

Adding $12$ to both sides gives $15r=32$ , so $r=\frac{32}{15}$ .

Question 45·Hard·Systems of Two Linear Equations in Two Variables

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Step-by-step Explanation

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Step-by-step Explanation