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Question 42·Hard·Systems of Two Linear Equations in Two Variables

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\begin{cases} (k+1)x + 2y = 7 \\ 3x + (k-2)y = 1 \end{cases}

In the system of equations above, $k$ is a constant. The system has a unique solution in which $x+y=4$ . What is the sum of all possible values of $k$ ?

Your answer

When a system includes a condition on its solution, such as $x+y=4$ , use that condition first to express one variable in terms of the other. Substitute into both equations to create two relationships involving the remaining variable and the parameter, then eliminate the variable to solve for the parameter. Finally, check any stated condition such as uniqueness; for a two-equation linear system, the variable coefficients must not be proportional.

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Use the condition involving x and y

The condition $x+y=4$ lets you replace $y$ with an expression involving $x$ . Substitute that expression into both equations.

Create two equations involving x and k

After substituting $y=4-x$ , simplify each equation. Each should relate $x$ to $k$ .

Eliminate x

Use one of the simplified equations to replace $x$ in the other. The result should be a quadratic equation in $k$ .

Check uniqueness

After finding the possible values of $k$ , verify that the variable coefficients are not proportional. Proportional coefficients would prevent the system from having a unique solution.

Desmos Guide

Derive the equation in k algebraically

First use $y=4-x$ in the two equations and eliminate $x$ . This produces the equation $k^2-3k+1=0$ .

Graph the quadratic

In Desmos, enter y=x^2-3x+1, using $x$ as a temporary variable representing $k$ . The $x$ -intercepts represent the possible values of $k$ .

Use symmetry to find the sum

Enter x=3/2 to display the axis of symmetry. The two intercepts are symmetric about this line, so their sum is twice the $x$ -coordinate of the axis.

Verify the unique-solution condition

For each intercept, evaluate x^2-x-8. A result other than 0 confirms that the variable coefficients are not proportional, so the system has a unique solution.

Step-by-step Explanation

Use the condition on the solution

Because $x+y=4$ , write

y=4-x.

Substitute this expression into each equation of the system.

Rewrite both equations using x and k

Substituting $y=4-x$ into the first equation gives

(k+1)x+2(4-x)=7,

(k-1)x=-1.

Substituting into the second equation gives

3x+(k-2)(4-x)=1,

(5-k)x=9-4k.

Eliminate x and solve for k

From $(k-1)x=-1$ , replace $x$ in the other equation. This gives

-\frac{5-k}{k-1}=9-4k.

Multiplying by $k-1$ and simplifying:

\begin{aligned} k-5&=(k-1)(9-4k)\\ k-5&=-4k^2+13k-9\\ k^2-3k+1&=0. \end{aligned}

Using the quadratic formula,

k=\frac{3\pm\sqrt{5}}{2}.

Confirm that each value gives a unique solution

A system of two linear equations does not have a unique solution only when its variable coefficients are proportional. Here, that would require

(k+1)(k-2)=6,

or equivalently

k^2-k-8=0.

For either candidate value, $k^2=3k-1$ , so the left side becomes

k^2-k-8=2k-9,

which is not $0$ for either value of $k$ . Therefore, both values produce a unique solution.

Add the possible values

The two possible values are

\frac{3+\sqrt{5}}{2} \quad\text{and}\quad \frac{3-\sqrt{5}}{2}.

Their sum is

\frac{3+\sqrt{5}}{2}+\frac{3-\sqrt{5}}{2}=3.

Question 42·Hard·Systems of Two Linear Equations in Two Variables

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation