The system of equations \(2x-3y-5 = 0\), \(15y-10x + 50 = 0\)
We are given a system of two linear equations:
To analyze the system, let's rewrite both equations in the standard form \(Ax + By = C\).
We can determine if a system of linear equations \(A_1x + B_1y = C_1\) and \(A_2x + B_2y = C_2\) has a unique solution, infinitely many solutions, or no solution (inconsistent) by comparing the ratios of their coefficients:
Let's calculate the ratios for the given system:
Now, let's compare these ratios:
We observe that \(\frac{A_1}{A_2} = -\frac{1}{5}\) and \(\frac{B_1}{B_2} = -\frac{1}{5}\). So, \(\frac{A_1}{A_2} = \frac{B_1}{B_2}\).
However, when we compare this to the ratio of the constant terms, we see that \(-\frac{1}{5} \neq -\frac{1}{10}\). Therefore, \(\frac{A_1}{A_2} = \frac{B_1}{B_2} \neq \frac{C_1}{C_2}\).
According to the conditions for analyzing systems of linear equations, the relationship \(\frac{A_1}{A_2} = \frac{B_1}{B_2} \neq \frac{C_1}{C_2}\) indicates that the system of equations is inconsistent. This means there are no values of \(x\) and \(y\) that can satisfy both equations simultaneously, as the lines represented by the equations are parallel.
If 2 x + 3 y = 17;
2 x+2 - 3 y+1 = 5
then the values of x and y are:
The solution of pair of linear equations \(\dfrac{1}{2}x+\dfrac{2}{3}y=-1,x-\dfrac{1}{3}y=3\) by the elimination method, is:
Kumar tried his skill at shooting at a fun fair. He has to hit the target and if he hits the target he gets 1 Rs. and if he misses he has to pay 50 paise. He attempted 25 shots and won 10 Rs. In how many did he hit the target?
The sum of two numbers is 66 and their difference is 22. What is the ratio of the two numbers?
Shyam spent half of his money and was left with as many as he had rupees before, but with half as many rupees as he had paise before. Which of the following is a possible amount of money he is left with?