Question

For what value(s) of k will the following system of linear equations have no solution? infinitely many solutions? x - 2y = 3 -2x + 4y = k

Accepted Answer

Write down the system of equations clearly: 
$$ x - 2y = 3 $$ 
$$ -2x + 4y = k $$ Recognize that for a system of two linear equations, the system will have:
- No solution if the lines are parallel but not the same line.
- Infinitely many solutions if the two equations represent the same line. To analyze this, compare the ratios of the coefficients of $$x$$, $$y$$, and the constants. For the system:
$$ \frac{a_1$}{a_2$} = \frac{b_1$}{b_2$} 
eq \frac{c_1$}{c_2$} $$ means no solution,
$$ \frac{a_1$}{a_2$} = \frac{b_1$}{b_2$} = \frac{c_1$}{c_2$} $$ means infinitely many solutions,
where $$a_1$$, $$b_1$$, $$c_1$$ are coefficients from the first equation and $$a_2$$, $$b_2$$, $$c_2$$ from the second. Identify the coefficients:
First equation: $$a_1 = 1$, b_1$ = -2$$, $$c_1 = 3$
Second equation: a_2$ = -2$$, $$b_2 = 4$, c_2$ = k$$ Set up the ratios and solve for $$k$$:
Calculate $$\frac{a_1$}{a_2$}$$ and $$\frac{b_1$}{b_2$}$$ and check if they are equal.
Then, for infinitely many solutions, set $$\frac{c_1$}{c_2$}$$ equal to these ratios and solve for $$k.
$$For no solution, $$\frac{a_1$}{a_2$} = \frac{b_1$}{b_2$} 
eq \frac{c_1$}{c_2$}$$, so find $$k$$ values that do not satisfy the equality with $$c_1$$/$$c_2.$$

For what value(s) of k will the following system of linear equations have no solution? infinitely many solutions?
x - 2y = 3
-2x + 4y = k

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Key Concepts

Systems of Linear Equations

Conditions for No Solution

Conditions for Infinitely Many Solutions