Question

Solve for x and y:
\[\dfrac{2}{3x+2y}+\dfrac{3}{3x-2y}=\dfrac{17}{5}\] ; \[\dfrac{5}{3x+2y}+\dfrac{1}{3x-2y}=2\]

Answer 1

Hint: Now here to solve this equation and find the value of x and y we need to convert it into a simplified linear equation with two variables and solve the equations to find the value. We can solve a linear equation with arithmetic operations like multiplying addition division and subtraction to simplify the equations and find the values of the linear equation.

Complete step-by-step solution:
The equations that are given to us through which we can find the values of x and y are
\[\dfrac{2}{3x+2y}+\dfrac{3}{3x-2y}=\dfrac{17}{5}\]
\[\dfrac{5}{3x+2y}+\dfrac{1}{3x-2y}=2\]
Now here we need to simplify these equations into simple linear equations of two variables therefore we substitute here that
\[\dfrac{1}{3x+2y}=a\] ; \[\dfrac{1}{3x-2y}=b\]
Now putting these substitutions in this questions given to us we get that
\[2a+3b=\dfrac{17}{5}\] -----equation I
\[5a+b=2\] -----equation II
Now to simplify we multiply both sides of equation I with 5 and get
\[5\left( 2a+3b \right)=5\times \dfrac{17}{5}\]
Opening bracket
\[10a+15b=17\] -----equation III
Now multiplying equation II with 2 on both sides we get
\[2\left( 5a+b \right)=2\times 2\]
Opening bracket
\[10a+2b=4\]-----equation IV
Now subtracting equation III and IV to simplify and combine both equations
\[10a+15b-10a-2b=17-4\]
Now subtracting the terms with the same variable and cancelling the same terms we get
\[13b=13\]
Dividing both sides with 13 we get the value of b which is
\[b=1\]
Now substituting this value of b in equation III to get the value of a
\[10a+15(1)=17\]
Multiplying and subtracting;
\[10a=17-15\]
Dividing both sides by 10 we get value of a
\[a=\dfrac{2}{10}\]
Simplifying we get
\[a=\dfrac{1}{5}\] ; \[b=1\]
Now putting these values in the values of a and b we assumed at the start that is
\[\dfrac{1}{3x+2y}=a\] ; \[\dfrac{1}{3x-2y}=b\]
\[\dfrac{1}{3x+2y}=\dfrac{1}{5}\] ; \[\dfrac{1}{3x-2y}=1\]
Cross multiplying in both equations we get
\[3x+2y=5\]; \[3x-2y=1\]
Now adding both these equations to get the value of x and y
\[3x+2y+3x-2y=5+1\]
Cancelling the common terms and adding the same values
\[6x=6\]
Dividing both sides by 6 to get the value of x
\[x=1\]
Now substituting this value of x to get the value of x in \[3x+2y=5\]
\[3(1)+2y=5\]
Multiplying and subtracting
\[2y=2\]
Dividing with 2 from both sides we get
\[y=1\]
Therefore the values for x and y for this question is \[x=1\] ; \[y=1\].

Note: We can also solve this by substitution method where in equation III we can but the value of b which is found from equation II to find the value of a and b
\[5a+b=2\] ---- equation II
\[b=2-5a\]
Now putting this value of b in equation III
\[10a+15(2-5a)=17\]
Now through this we can find values of a and b and hence find x,y values too.