We are given two equations:
First, let's simplify the second equation:
Cross-multiply the terms:
$ 6x = 5(x+y) $
Distribute the 5:
$ 6x = 5x + 5y $
Subtract $5x$ from both sides to express $x$ in terms of $y$:
$ 6x - 5x = 5y $
$ x = 5y $
Now, substitute the expression for $x$ (from the simplified second equation) into the first equation:
Substitute $x = 5y$ into $2x - 3y = 7$:
$ 2(5y) - 3y = 7 $
Simplify the equation:
$ 10y - 3y = 7 $
$ 7y = 7 $
Solve for $y$:
$ y = \frac{7}{7} = 1 $
Now, use the relation $x = 5y$ to find the value of $x$:
$ x = 5(1) = 5 $
Finally, calculate the value of $x - y$ using the values we found for $x$ and $y$:
$ x - y = 5 - 1 $
$ x - y = 4 $
Therefore, the value of $x - y$ is 4.
If $2x - 3y = -1$ and $\frac{x}{x+y} = \frac{7}{12}$, then the value of $2xy$ is:
What is the solution of the following equations ?
2x + 3y = 12 and 3x − 2y = 5
Two positive numbers differ by 1280. When the greater number is divided by the smaller number, the quotient is 7 and the remainder is 50. The greater number is:
When 5 children from class A join class B, the number of children in both classes is the same. If 25 children from B, join A, then the number of children in A becomes double the number of children in B. The ratio of the number of children in A to those in B is:
If (x + 6y) = 8, and xy = 2, where x > 0, what is the value of (x 3+ 216y 3)?
If 8k 6+ 15k 3– 2 = 0, then the positive value of \(\left( {{\rm{k}}\,{\rm{ + }}\,\frac{1}{{\rm{k}}}} \right)\) is :