Answer
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Hint: Here, we are required to find the value of \[x - y\]. We would solve the given equations by any of the accepted methods and find the values of \[x\] and \[y\] respectively. Substituting their values in\[x - y\], we would get our required answer.
Complete step-by-step answer:
The given two equations are:
\[19x - 17y = 55\] ………..\[\left( 1 \right)\]
\[17x - 19y = 53\]………..\[\left( 2 \right)\]
We will multiply equation \[\left( 1 \right)\] by 17, we get
\[\left( {19x - 17y} \right) \times 17 = 55 \times 17\]
\[ \Rightarrow \left( {19 \times 17} \right)x - {17^2}y = 55 \times 17\]……………………….\[\left( 3 \right)\]
We will multiply equation \[\left( 2 \right)\] by 19, we get
\[\left( {17x - 19y} \right) \times 19 = 53 \times 19\]
\[ \Rightarrow \left( {17 \times 19} \right)x \times {19^2}y = 53 \times 19\]………………………\[\left( 4 \right)\]
Hence, subtracting the equations, we get,
Hence, subtracting the equation \[\left( 4 \right)\] from equation \[\left( 3 \right)\], we get,
\[\left[ {\left( {19 \times 17} \right)x - {{17}^2}y} \right] - \left[ {\left( {19 \times 17} \right)x - {{19}^2}y} \right] = 55 \times 17 - 53 \times 19\]
\[ \Rightarrow \left( {19 \times 17} \right)x - {17^2}y - \left( {19 \times 17} \right)x + {19^2}y = 55 \times 17 - 53 \times 19\]
Solving the terms with same variables, we get,
\[ \Rightarrow \left( {{{19}^2} - {{17}^2}} \right)y = \left( {55 \times 17} \right) - \left( {53 \times 19} \right)\]
Now, applying the identity \[\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)\]
\[ \Rightarrow \left( {19 + 17} \right)\left( {19 - 17} \right)y = \left( {55 \times 17} \right) - \left( {53 \times 19} \right)\]
Now, we would transport everything which is multiplying to the variable \[y\] in the LHS to the denominator of RHS.
\[ \Rightarrow y = \dfrac{{\left( {55 \times 17} \right) - \left( {53 \times 19} \right)}}{{\left( {19 + 17} \right)\left( {19 - 17} \right)}}\]
Now, we would solve this further,
\[ \Rightarrow y = \dfrac{{935 - 1007}}{{36 \times 2}}\]
\[ \Rightarrow y = \dfrac{{ - 72}}{{72}}\]
\[ \Rightarrow y = - 1\]
Now, substituting this value in the equation,\[19x - 17y = 55\], we get,
\[19x - 17\left( { - 1} \right) = 55\]
\[19x + 17 = 55\]
Subtracting 17 from both sides,
\[19x = 38\]
Dividing both sides by 19
\[ \Rightarrow x = 2\]
Therefore, the required value of \[x - y\] is:
\[x - y = 2 - \left( { - 1} \right)\]
\[ \Rightarrow x - y = 2 + 1 = 3\]
Therefore, if \[19x - 17y = 55\] and \[17x - 19y = 53\], then the value of \[x - y = 3\]
Hence, option C is the required answer.
Note:
We can notice in the given two equations that the coefficient of \[x\] in the first equation is the coefficient of \[y\] in the second equation. Similarly, the coefficient of \[y\] in the first equation is the coefficient of \[x\] in the second equation.
Hence, there is another way to solve this question.
The given two equations are:
\[19x - 17y = 55\] ………..\[\left( 1 \right)\]
\[17x - 19y = 53\]………..\[\left( 2 \right)\]
Now, adding the equations \[\left( 1 \right)\] and \[\left( 2 \right)\], we get,
\[19x - 17y + 17x - 19y = 55 + 53\]
\[ \Rightarrow 36x - 36y = 108\]
Now, taking 36 common in the LHS,
\[ \Rightarrow 36\left( {x - y} \right) = 108\]
Dividing both sides by 36
\[ \Rightarrow x - y = 3\]…………………………\[\left( 5 \right)\]
Similarly, subtracting the equations \[\left( 1 \right)\] and \[\left( 2 \right)\], we get,
\[19x - 17y - 17x + 19y = 55 - 53\]
\[ \Rightarrow 2x + 2y = 2\]
Now, taking 2 common in the LHS,
\[ \Rightarrow 2\left( {x + y} \right) = 2\]
Dividing both sides by 2
\[ \Rightarrow x + y = 1\]…………………………\[\left( 6 \right)\]
Now, adding the equations \[\left( 5 \right)\] and \[\left( 6 \right)\] using elimination method,
\[ \Rightarrow x - y + x + y = 3 + 1\]
\[ \Rightarrow 2x = 4\]
Dividing both sides by 2, we get
\[ \Rightarrow x = 2\]
Substituting this value in equation \[\left( 5 \right)\]
\[ \Rightarrow x - y = 3\]
\[ \Rightarrow 2 - y = 3\]
\[ \Rightarrow y = 2 - 3 = - 1\]
Therefore, the required value of \[x - y\] is:
\[x - y = 2 - \left( { - 1} \right)\]
\[ \Rightarrow x - y = 2 + 1 = 3\]
Hence, we could attempt this question by either of the two ways.
Complete step-by-step answer:
The given two equations are:
\[19x - 17y = 55\] ………..\[\left( 1 \right)\]
\[17x - 19y = 53\]………..\[\left( 2 \right)\]
We will multiply equation \[\left( 1 \right)\] by 17, we get
\[\left( {19x - 17y} \right) \times 17 = 55 \times 17\]
\[ \Rightarrow \left( {19 \times 17} \right)x - {17^2}y = 55 \times 17\]……………………….\[\left( 3 \right)\]
We will multiply equation \[\left( 2 \right)\] by 19, we get
\[\left( {17x - 19y} \right) \times 19 = 53 \times 19\]
\[ \Rightarrow \left( {17 \times 19} \right)x \times {19^2}y = 53 \times 19\]………………………\[\left( 4 \right)\]
Hence, subtracting the equations, we get,
Hence, subtracting the equation \[\left( 4 \right)\] from equation \[\left( 3 \right)\], we get,
\[\left[ {\left( {19 \times 17} \right)x - {{17}^2}y} \right] - \left[ {\left( {19 \times 17} \right)x - {{19}^2}y} \right] = 55 \times 17 - 53 \times 19\]
\[ \Rightarrow \left( {19 \times 17} \right)x - {17^2}y - \left( {19 \times 17} \right)x + {19^2}y = 55 \times 17 - 53 \times 19\]
Solving the terms with same variables, we get,
\[ \Rightarrow \left( {{{19}^2} - {{17}^2}} \right)y = \left( {55 \times 17} \right) - \left( {53 \times 19} \right)\]
Now, applying the identity \[\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)\]
\[ \Rightarrow \left( {19 + 17} \right)\left( {19 - 17} \right)y = \left( {55 \times 17} \right) - \left( {53 \times 19} \right)\]
Now, we would transport everything which is multiplying to the variable \[y\] in the LHS to the denominator of RHS.
\[ \Rightarrow y = \dfrac{{\left( {55 \times 17} \right) - \left( {53 \times 19} \right)}}{{\left( {19 + 17} \right)\left( {19 - 17} \right)}}\]
Now, we would solve this further,
\[ \Rightarrow y = \dfrac{{935 - 1007}}{{36 \times 2}}\]
\[ \Rightarrow y = \dfrac{{ - 72}}{{72}}\]
\[ \Rightarrow y = - 1\]
Now, substituting this value in the equation,\[19x - 17y = 55\], we get,
\[19x - 17\left( { - 1} \right) = 55\]
\[19x + 17 = 55\]
Subtracting 17 from both sides,
\[19x = 38\]
Dividing both sides by 19
\[ \Rightarrow x = 2\]
Therefore, the required value of \[x - y\] is:
\[x - y = 2 - \left( { - 1} \right)\]
\[ \Rightarrow x - y = 2 + 1 = 3\]
Therefore, if \[19x - 17y = 55\] and \[17x - 19y = 53\], then the value of \[x - y = 3\]
Hence, option C is the required answer.
Note:
We can notice in the given two equations that the coefficient of \[x\] in the first equation is the coefficient of \[y\] in the second equation. Similarly, the coefficient of \[y\] in the first equation is the coefficient of \[x\] in the second equation.
Hence, there is another way to solve this question.
The given two equations are:
\[19x - 17y = 55\] ………..\[\left( 1 \right)\]
\[17x - 19y = 53\]………..\[\left( 2 \right)\]
Now, adding the equations \[\left( 1 \right)\] and \[\left( 2 \right)\], we get,
\[19x - 17y + 17x - 19y = 55 + 53\]
\[ \Rightarrow 36x - 36y = 108\]
Now, taking 36 common in the LHS,
\[ \Rightarrow 36\left( {x - y} \right) = 108\]
Dividing both sides by 36
\[ \Rightarrow x - y = 3\]…………………………\[\left( 5 \right)\]
Similarly, subtracting the equations \[\left( 1 \right)\] and \[\left( 2 \right)\], we get,
\[19x - 17y - 17x + 19y = 55 - 53\]
\[ \Rightarrow 2x + 2y = 2\]
Now, taking 2 common in the LHS,
\[ \Rightarrow 2\left( {x + y} \right) = 2\]
Dividing both sides by 2
\[ \Rightarrow x + y = 1\]…………………………\[\left( 6 \right)\]
Now, adding the equations \[\left( 5 \right)\] and \[\left( 6 \right)\] using elimination method,
\[ \Rightarrow x - y + x + y = 3 + 1\]
\[ \Rightarrow 2x = 4\]
Dividing both sides by 2, we get
\[ \Rightarrow x = 2\]
Substituting this value in equation \[\left( 5 \right)\]
\[ \Rightarrow x - y = 3\]
\[ \Rightarrow 2 - y = 3\]
\[ \Rightarrow y = 2 - 3 = - 1\]
Therefore, the required value of \[x - y\] is:
\[x - y = 2 - \left( { - 1} \right)\]
\[ \Rightarrow x - y = 2 + 1 = 3\]
Hence, we could attempt this question by either of the two ways.
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