For any 2 point A(0,0) , B(
x
o
,
y
o
{\displaystyle x_{o},y_{o}}
m
=
y
o
x
o
{\displaystyle m={\frac {y_{o}}{x_{o}}}}
Therefore,
y
o
=
m
x
o
{\displaystyle y_{o}=mx_{o}}
For a point C(
x
,
y
{\displaystyle x,y}
m
=
y
−
y
o
x
−
x
o
{\displaystyle m={\frac {y-y_{o}}{x-x_{o}}}}
y
=
y
o
+
m
(
x
−
x
o
)
{\displaystyle y=y_{o}+m(x-x_{o})}
Has general form
A
x
=
0
{\displaystyle Ax=0}
x
=
0
{\displaystyle x=0}
A
x
=
C
{\displaystyle Ax=C}
x
=
C
A
{\displaystyle x={\frac {C}{A}}}
A
x
+
B
=
C
{\displaystyle Ax+B=C}
x
=
C
−
B
A
{\displaystyle x={\frac {C-B}{A}}}
A
x
+
B
y
=
C
{\displaystyle Ax+By=C}
x
=
C
A
{\displaystyle x={\frac {C}{A}}}
y
=
0
{\displaystyle y=0}
y
=
C
B
{\displaystyle y={\frac {C}{B}}}
x
=
0
{\displaystyle x=0}
2
x
+
y
=
11
{\displaystyle 2x+y=11\,}
−
4
x
+
3
y
=
13
{\displaystyle -4x+3y=13\,}
Roots of system of equations [ edit | edit source ]
2
x
+
y
=
11
{\displaystyle 2x+y=11\,}
−
4
x
+
3
y
=
13
{\displaystyle -4x+3y=13\,}
Multiply 1st row by 2 add to 2nd row
4
x
+
2
y
=
22
{\displaystyle 4x+2y=22\,}
−
4
x
+
3
y
=
13
{\displaystyle -4x+3y=13\,}
Yields
5
y
=
35
{\displaystyle 5y=35\,}
y
=
7
{\displaystyle y=7}
Multiply 1st row by -3 add to 2nd row
−
6
x
−
3
y
=
−
33
{\displaystyle -6x-3y=-33\,}
−
4
x
+
3
y
=
13
{\displaystyle -4x+3y=13\,}
Yields
−
10
x
=
−
20
{\displaystyle -10x=-20\,}
x
=
2
{\displaystyle x=2}
If you get a system of equations that looks like this:
2
x
+
y
=
11
{\displaystyle 2x+y=11\,}
−
4
x
+
3
y
=
13
{\displaystyle -4x+3y=13\,}
You can switch around some terms in the first to get this:
y
=
−
2
x
+
11
{\displaystyle y=-2x+11\,}
Then you can substitute that into the bottom one so that it looks like this:
−
4
x
+
3
(
−
2
x
+
11
)
=
13
{\displaystyle -4x+3(-2x+11)=13\,}
−
4
x
−
6
x
+
33
=
13
{\displaystyle -4x-6x+33=13\,}
−
10
x
+
33
=
13
{\displaystyle -10x+33=13\,}
−
10
x
=
−
20
{\displaystyle -10x=-20\,}
x
=
2
{\displaystyle x=2\,}
Then, you can substitute 2 into an x from either equation and solve for y . It's usually easier to substitute it in the one that had the single y . In this case, after substituting 2 for x , you would find that y = 7.
If you get a system of equations that looks like this:
2
x
+
y
=
11
{\displaystyle 2x+y=11\,}
−
4
x
+
3
y
=
13
{\displaystyle -4x+3y=13\,}
Solve for y
x
+
1
2
y
=
11
2
{\displaystyle x+{\frac {1}{2}}y={\frac {11}{2}}\,}
x
+
3
−
4
y
=
13
−
4
{\displaystyle x+{\frac {3}{-4}}y={\frac {13}{-4}}\,}
y
(
1
2
−
3
−
4
)
=
11
2
−
13
−
4
{\displaystyle y({\frac {1}{2}}-{\frac {3}{-4}})={\frac {11}{2}}-{\frac {13}{-4}}}
y
=
11
2
−
13
−
4
1
2
−
3
−
4
{\displaystyle y={\frac {{\frac {11}{2}}-{\frac {13}{-4}}}{{\frac {1}{2}}-{\frac {3}{-4}}}}}
Solve for x
2
x
+
y
=
11
{\displaystyle 2x+y=11\,}
−
4
3
x
+
y
=
13
3
{\displaystyle {\frac {-4}{3}}x+y={\frac {13}{3}}\,}
x
(
2
−
−
4
3
)
=
11
−
13
3
{\displaystyle x(2-{\frac {-4}{3}})=11-{\frac {13}{3}}}
x
=
11
−
13
3
2
−
−
4
3
{\displaystyle x={\frac {11-{\frac {13}{3}}}{2-{\frac {-4}{3}}}}}
Find Roots of System of Linear Equations systematically [ edit | edit source ]
A
1
x
+
B
1
y
=
C
1
{\displaystyle A_{1}x+B_{1}y=C_{1}}
A
2
x
+
B
2
y
=
C
2
{\displaystyle A_{2}x+B_{2}y=C_{2}}
Eliminate variable x
x
+
B
1
A
1
y
=
C
1
A
1
{\displaystyle x+{\frac {B_{1}}{A_{1}}}y={\frac {C_{1}}{A_{1}}}}
x
+
B
2
A
2
y
=
C
2
A
2
{\displaystyle x+{\frac {B_{2}}{A_{2}}}y={\frac {C_{2}}{A_{2}}}}
Add
y
(
B
1
A
1
−
B
2
A
2
)
=
C
1
A
1
−
C
2
A
2
{\displaystyle y({\frac {B_{1}}{A_{1}}}-{\frac {B_{2}}{A_{2}}})={\frac {C_{1}}{A_{1}}}-{\frac {C_{2}}{A_{2}}}}
Solve for y
y
=
C
1
A
1
−
C
2
A
2
B
1
A
1
−
B
2
A
2
{\displaystyle y={\frac {{\frac {C_{1}}{A_{1}}}-{\frac {C_{2}}{A_{2}}}}{{\frac {B_{1}}{A_{1}}}-{\frac {B_{2}}{A_{2}}}}}}
Eliminate variable y
A
1
B
1
x
+
y
=
C
1
B
1
{\displaystyle {\frac {A_{1}}{B_{1}}}x+y={\frac {C_{1}}{B_{1}}}}
A
2
B
2
x
+
y
=
C
2
B
2
{\displaystyle {\frac {A_{2}}{B_{2}}}x+y={\frac {C_{2}}{B_{2}}}}
Add
x
(
A
1
B
1
−
A
2
B
2
)
=
C
1
B
1
−
C
2
B
2
{\displaystyle x({\frac {A_{1}}{B_{1}}}-{\frac {A_{2}}{B_{2}}})={\frac {C_{1}}{B_{1}}}-{\frac {C_{2}}{B_{2}}}}
Solve for x
x
=
C
1
B
1
−
C
2
B
2
A
1
B
1
−
A
2
B
2
{\displaystyle x={\frac {{\frac {C_{1}}{B_{1}}}-{\frac {C_{2}}{B_{2}}}}{{\frac {A_{1}}{B_{1}}}-{\frac {A_{2}}{B_{2}}}}}}
