Question

Solve. \[\dfrac{{7 + x}}{5} - \dfrac{{2x - y}}{4} = 3y - 5\] and \[\dfrac{{5y - 7}}{2} + \dfrac{{4x - 3}}{6} = 18 - 5x\].

Answer 1

Hint: First, we simplify the given equations, so that we can solve them easily. Then we use the elimination method, which states that the coefficient of x in equation 1 is multiplied with equation 2 and at the same time, coefficient of x in equation 1 is multiplied with equation 2. Then we change the sign of equation 2 and add it in equation 1. So, we get the value of y. Now, we substitute the value of y in any equation to get the value of x.

Complete step-by-step answer:
First of all, we will simplify the equation \[\dfrac{{7 + x}}{5} - \dfrac{{2x - y}}{4} = 3y - 5\].
Multiplying both sides by \[20\], we get,
\[ \Rightarrow 20\left( {\dfrac{{7 + x}}{5} - \dfrac{{2x - y}}{4}} \right) = 20\left( {3y - 5} \right)\]
\[ \Rightarrow 20\left( {\dfrac{{7 + x}}{5}} \right) - 20\left( {\dfrac{{2x - y}}{4}} \right) = 20\left( {3y - 5} \right)\]
Opening the brackets and simplifying the equation, we get,
\[ \Rightarrow 4\left( {7 + x} \right) - 5\left( {2x - y} \right) = 20\left( {3y - 5} \right)\]
\[ \Rightarrow 28 + 4x - 10x + 5y = 60y - 100\]
Now, taking all terms to right hand side we get,
\[ \Rightarrow 0 = 60y - 100 - 28 - 4x + 10x - 5y\]
On simplifying, we get,
\[ \Rightarrow 0 = 6x + 55y - 128\]
Taking constant term to left hand side,
\[ \Rightarrow 128 = 6x + 55y\]
This can also be written as
\[ \Rightarrow 6x + 55y = 128\]
Now, we will simplify equation \[\dfrac{{5y - 7}}{2} + \dfrac{{4x - 3}}{6} = 18 - 5x\]
Multiplying both sides by \[6\], we get,
\[ \Rightarrow 6\left( {\dfrac{{5y - 7}}{2} + \dfrac{{4x - 3}}{6}} \right) = 6\left( {18 - 5x} \right)\]
\[ \Rightarrow 6\left( {\dfrac{{5y - 7}}{2}} \right) + 6\left( {\dfrac{{4x - 3}}{6}} \right) = 6\left( {18 - 5x} \right)\]
Simplifying the equation,
\[ \Rightarrow 3\left( {5y - 7} \right) + \left( {4x - 3} \right) = 6\left( {18 - 5x} \right)\]
\[ \Rightarrow 15y - 21 + 4x - 3 = 108 - 30x\]
Separating terms having x and y in left side and constant term in right side we get,
\[ \Rightarrow 15y + 4x + 30x = 108 + 21 + 3\]
\[ \Rightarrow 34x + 15y = 132\]
On simplifying both equation we get,
\[6x + 55y = 128\] _ _ _ _ _ _ _ _ (1)
\[34x + 15y = 132\] _ _ _ _ _ _ _ _ _ _ (2)
Now, using elimination method, that is,
Multiplying equation \[\left( 2 \right)\] with \[6\] and equation \[\left( 1 \right)\] with\[34\], we get
\[ \Rightarrow 34\left( {6x + 55y = 128} \right)\]
\[204x + 1870y = 4352\] _ _ _ (a)
\[ \Rightarrow 6\left( {34x + 15y = 132} \right)\]
\[204x + 90y = 792\] _ _ _ _ (b)
Subtracting equation (b) from equation (a), we get,
\[ \Rightarrow 204x + 1870y - \left( {204x + 90y} \right) = 4352 - 792\]
\[ \Rightarrow 204x + 1870y - 204x - 90y = 3560\]
We get,
\[ \Rightarrow 1870y - 90y = 3560\]
\[ \Rightarrow 1780y = 3560\]
Dividing both sides by \[1780\] , we get,
\[ \Rightarrow y = \dfrac{{3560}}{{1780}} = 2\]
Now, substituting value of y in equation (1)
We get,
\[ \Rightarrow 6x + 55\left( 2 \right) = 128\]
\[ \Rightarrow 6x + 110 = 128\]
Simplifying the equation, we get,
\[ \Rightarrow 6x = 18\]
Dividing both sides by six, we get,
\[ \Rightarrow x = 3\]
Hence, we get \[x = 3\] and \[y = 2\].

Note: This question deals with the understanding of different methods of solving the given set of equations. We have used an elimination method. We can also use the substitution method and cross multiplication method to solve a given set of equations. Take care while doing the calculation so as to be sure of the final answer.