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Question
Which of the following systems of equations have (−5,−4) as a solution?
System PSystem QSystem RSystem Sy=−2x−14x−5y=153x+3y=−27y+2x=−14y=2x+14y=23x−23x−y=−1x−2y=3
System PSystem QSystem RSystem Sy=−2x−14x−5y=153x+3y=−27y+2x=−14y=2x+14y=23x−23x−y=−1x−2y=3
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Solution
Detailed step-by-step solution:
Solving the system P of equations:
y=−2x−14 ...equation (1)
y=2x+14 ...equation (2)
Adding equation (1) and equation (2), we get:
y=−2x−14
+y=2x+14
2y=0+0
⇒2y=0
⇒y=0
Putting y=0 in equation (1),
0=−2x−14
2x=−14
⇒x=−142 (dividing both sides by 2)
⇒x=−7
Solution is (−7,0).
System P does not have (−5,−4) as a solution.
Solving the system Q of equations:
x−5y=15 ...equation (1)
y=23x−23 ...equation (2)
Multiply equation (2) with 3 to eliminate the fraction from the equation
3(y=23x−23).
⇒3y=2x−2
⇒−2x+3y=−2 ...equation (3)
To make the coefficient of x the same, multiply equation (1) with 2:
2(x−5y=15)
⇒2x−10y=30 ...equation (4)
Adding equations (3) and (4):
2x−10y=30
−2x+3y=−2
0−7y=28 [adding equations (3) and 4]
⇒−7y=28
⇒y=−4,
Substitute y=−4 in equation (1):
⇒x−5(−4)=15
⇒x+20=15
⇒x=15−20
⇒x=−5
Solution is (−5,−4).
Solving system R:
3x+3y=−27 ...equation (1)
x−y=−1 ...equation (2)
Multiply equation (2) with 3 to get the same coefficients of x and y, we get:
3x−3y=−3 ...equation (3)
Adding equation (1) and equation (3), we get:
3x+3y=−27
3x−3y=−3
6x+0=−30
⇒x=−306
⇒x=−5
Putting x=−5 in equation (2), we get:
−5−y=−1
⇒−y=−1+5
⇒−y=4
⇒y=−4
Solution is (−5,−4).
Solving system S:
y+2x=−14 ...equation (1)
x−2y=3 ...
−2y+x=3 ...equation (2) (reversing above equation)
Multiplying equation 2 with 2 to get the coefficient of x to be the same, we get:
−4y+2x=6 ...equation (3)
Subtracting equation (3) from equation (1):
y+2x=−14
−4y+2x=6
5y+0=−20
⇒5y=−20
⇒y=−205
⇒y=−4
Putting y=−4 in equation (1), we get:
−4+2x=−14
⇒2x=−14+4
⇒2x=−10
⇒x=−5
Solution is (−5,−4).
All the system has (−5,−4) as a solution except system P.
⇒ Option A is correct.
Solving the system P of equations:
y=−2x−14 ...equation (1)
y=2x+14 ...equation (2)
Adding equation (1) and equation (2), we get:
y=−2x−14
+y=2x+14
2y=0+0
⇒2y=0
⇒y=0
Putting y=0 in equation (1),
0=−2x−14
2x=−14
⇒x=−142 (dividing both sides by 2)
⇒x=−7
Solution is (−7,0).
System P does not have (−5,−4) as a solution.
Solving the system Q of equations:
x−5y=15 ...equation (1)
y=23x−23 ...equation (2)
Multiply equation (2) with 3 to eliminate the fraction from the equation
3(y=23x−23).
⇒3y=2x−2
⇒−2x+3y=−2 ...equation (3)
To make the coefficient of x the same, multiply equation (1) with 2:
2(x−5y=15)
⇒2x−10y=30 ...equation (4)
Adding equations (3) and (4):
2x−10y=30
−2x+3y=−2
0−7y=28 [adding equations (3) and 4]
⇒−7y=28
⇒y=−4,
Substitute y=−4 in equation (1):
⇒x−5(−4)=15
⇒x+20=15
⇒x=15−20
⇒x=−5
Solution is (−5,−4).
Solving system R:
3x+3y=−27 ...equation (1)
x−y=−1 ...equation (2)
Multiply equation (2) with 3 to get the same coefficients of x and y, we get:
3x−3y=−3 ...equation (3)
Adding equation (1) and equation (3), we get:
3x+3y=−27
3x−3y=−3
6x+0=−30
⇒x=−306
⇒x=−5
Putting x=−5 in equation (2), we get:
−5−y=−1
⇒−y=−1+5
⇒−y=4
⇒y=−4
Solution is (−5,−4).
Solving system S:
y+2x=−14 ...equation (1)
x−2y=3 ...
−2y+x=3 ...equation (2) (reversing above equation)
Multiplying equation 2 with 2 to get the coefficient of x to be the same, we get:
−4y+2x=6 ...equation (3)
Subtracting equation (3) from equation (1):
y+2x=−14
−4y+2x=6
5y+0=−20
⇒5y=−20
⇒y=−205
⇒y=−4
Putting y=−4 in equation (1), we get:
−4+2x=−14
⇒2x=−14+4
⇒2x=−10
⇒x=−5
Solution is (−5,−4).
All the system has (−5,−4) as a solution except system P.
⇒ Option A is correct.
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