Which system of equationds has a graph that shows intersecting lines with (3,-1) as solution?
Answers
Answer:
A system of equations.
To find:
The set of equations that shows intersecting lines on a graph.
Solution:
Consider a set of linear equations:
a_{1}x+b_{1}y+c_{1}=0a
1
x+b
1
y+c
1
=0
a_{2}x+b_{2}y+c_{2}=0a
2
x+b
2
y+c
2
=0
These equations of lines intersect at a point on a graph when,
\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}
a
2
a
1
=
b
2
b
1
Then they are said to be consistent and have a unique solution of the system of linear equations in two variables.
Consider the equations:
2x+4y=142x+4y=14
x+2y=7x+2y=7
Here, a_{1}=2,b_{1}=4,c_{1}=-14a
1
=2,b
1
=4,c
1
=−14
a_{2}=1,b_{2}=2,c_{2}=-7a
2
=1,b
2
=2,c
2
=−7
\frac{a_{1}}{a_{2}}=\frac{2}{1}
a
2
a
1
=
1
2
\frac{b_{1}}{b_{2}} =\frac{4}{2}=\frac{2}{1}
b
2
b
1
=
2
4
=
1
2
Here, \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}
a
2
a
1
=
b
2
b
1
.
Hence, these lines of equations do not intersect at a point. So, option (A) is not correct.
Consider the equations,
-3x+y=5−3x+y=5
6x-2y=16x−2y=1
Here, a_{1}=-3,b_{1}=1,c_{1}=-5a
1
=−3,b
1
=1,c
1
=−5
a_{2}=6,b_{2}=-2,c_{2}=-1a
2
=6,b
2
=−2,c
2
=−1
\frac{a_{1}}{a_{2}} =\frac{-3}{6}=\frac{-1}{2}
a
2
a
1
=
6
−3
=
2
−1
\frac{b_{1}}{b_{2}}=\frac{1}{-2}=\frac{-1}{2}
b
2
b
1
=
−2
1
=
2
−1
Here, \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}
a
2
a
1
=
b
2
b
1
.
Hence, these lines of equations do not intersect at a point. So, option (B) is not correct.
Consider the equations,
4x+8y=74x+8y=7
x+2y=3x+2y=3
Here, a_{1}=4,b_{1}=8,c_{1}=-7a
1
=4,b
1
=8,c
1
=−7
a_{2}=1,b_{2}=2,c_{2}=-3a
2
=1,b
2
=2,c
2
=−3
\frac{a_{1}}{a_{2}} =\frac{4}{1}
a
2
a
1
=
1
4
\frac{b_{1}}{b_{2}}=\frac{8}{2}=\frac{4}{1}
b
2
b
1
=
2
8
=
1
4
Here, \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}
a
2
a
1
=
b
2
b
1
.
Hence, these lines of equations do not intersect at a point. So, option (C) is not correct.
Consider the equations,
3x+y=103x+y=10
3x-y=53x−y=5
Here, a_{1}=3,b_{1}=1,c_{1}=-10a
1
=3,b
1
=1,c
1
=−10
a_{2}=3,b_{2}=-1,c_{2}=-5a
2
=3,b
2
=−1,c
2
=−5
\frac{a_{1}}{a_{2}} =\frac{3}{3}=\frac{1}{1}
a
2
a
1
=
3
3
=
1
1
\frac{b_{1}}{b_{2}}=\frac{1}{-1}=\frac{-1}{1}
b
2
b
1
=
−1
1
=
1
−1
Here, \frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}
a
2
a
1
=
b
2
b
1
.
Hence, these lines of equations intersect at a point A(2.5, 2.5) as shown in the figure below. So, option (D) is correct.
The lines of equations 3x+y=103x+y=10 and 3x-y=53x−y=5 intersect at a point A(2.5, 2.5) and is shown on a graph. Option (D) is the correct answer.