If the equations 6x – 5y + 11 = 0 and 15x + ky – 9 = 0 have no solution, then the value of k is:
-12.5
The problem asks us to find the value of 'k' for which the given system of linear equations has no solution. We are given two linear equations in two variables, x and y:
A system of two linear equations of the form \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\) is said to have no solution if the lines represented by these equations are parallel and distinct. This condition is met when the ratio of the coefficients of x is equal to the ratio of the coefficients of y, but this ratio is not equal to the ratio of the constant terms. Mathematically, this is expressed as:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \]From the given equations, we can identify the coefficients:
For equation 1, \(6x - 5y + 11 = 0\):
For equation 2, \(15x + ky - 9 = 0\):
Using the condition for no solution, \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\), we substitute the coefficients:
\[ \frac{6}{15} = \frac{-5}{k} \neq \frac{11}{-9} \]To find the value of k, we use the equality part of the condition:
\[ \frac{6}{15} = \frac{-5}{k} \]First, simplify the fraction \(\frac{6}{15}\). Both 6 and 15 are divisible by 3:
\[ \frac{6 \div 3}{15 \div 3} = \frac{2}{5} \]So, the equation becomes:
\[ \frac{2}{5} = \frac{-5}{k} \]Now, we can solve for k by cross-multiplication:
\[ 2 \times k = 5 \times (-5) \] \[ 2k = -25 \]Divide both sides by 2 to find k:
\[ k = \frac{-25}{2} \] \[ k = -12.5 \]We must also check if the inequality \(\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\) holds for \(k = -12.5\). This means checking if \(\frac{-5}{-12.5} \neq \frac{11}{-9}\).
The first ratio is \(\frac{-5}{-12.5} = \frac{5}{12.5}\). To remove the decimal, multiply numerator and denominator by 10:
\[ \frac{5 \times 10}{12.5 \times 10} = \frac{50}{125} \]Both 50 and 125 are divisible by 25:
\[ \frac{50 \div 25}{125 \div 25} = \frac{2}{5} \]The second ratio is \(\frac{11}{-9} = -\frac{11}{9}\).
We need to check if \(\frac{2}{5} \neq -\frac{11}{9}\). Since \(\frac{2}{5}\) is a positive value and \(-\frac{11}{9}\) is a negative value, they are indeed not equal. Thus, the condition \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\) is satisfied for \(k = -12.5\).
Therefore, the value of k for which the system of equations \(6x - 5y + 11 = 0\) and \(15x + ky - 9 = 0\) has no solution is -12.5.
| Condition | Geometric Interpretation | Algebraic Condition | Number of Solutions |
|---|---|---|---|
| Consistent and Unique Solution | Intersecting lines | \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\) | Exactly one solution |
| Consistent and Infinitely Many Solutions | Coincident lines | \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\) | Infinitely many solutions |
| Inconsistent / No Solution | Parallel and distinct lines | \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\) | No solution |
A system of linear equations that has no solution is called an inconsistent system. This means there is no pair of values (x, y) that can satisfy both equations simultaneously. When graphed, these equations represent lines that are parallel but do not overlap at any point. The distance between the two lines is constant, and they never intersect. The condition \(\frac{a_1}{a_2} = \frac{b_1}{b_2}\) ensures the lines are parallel (same slope), and the condition \(\neq \frac{c_1}{c_2}\) ensures they are distinct (different y-intercepts if \(b_1, b_2 \neq 0\)). Understanding these conditions is crucial for analyzing the nature of solutions for any system of two linear equations in two variables.
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