Here are the generic simultaneous equations.
-
ax
+
by
=
e
ax
+
by
=
e
size 12{ ital "ax"+ ital "by"=e} {}
-
cx
+
dy
=
f
cx
+
dy
=
f
size 12{ ital "cx"+ ital "dy"=f} {}
I call them “generic” because every possible pair of simultaneous equations looks exactly like that, except with numbers instead of
aa size 12{a} {},
bb size 12{b} {},
cc size 12{c} {},
dd size 12{d} {},
ee size 12{e} {}, and
ff size 12{f} {}. We are going to solve these equations.
When I say “solve it” I mean find a formula
x=x= size 12{x={}} {}blah-blah where the blah-blah has only
aa size 12{a} {},
bb size 12{b} {},
cc size 12{c} {},
dd size 12{d} {},
ee size 12{e} {}, and
ff size 12{f} {}: no
xx size 12{x} {} or
yy size 12{y} {}. And, of course,
y=y= size 12{y={}} {} some different formula with only
aa size 12{a} {},
bb size 12{b} {},
cc size 12{c} {},
dd size 12{d} {},
ee size 12{e} {}, and
ff size 12{f} {}. If we can do that, we will be able to use these formulas to immediately solve any pair of simultaneous equations, just by plugging in the numbers.
We can solve this by elimination or by substitution. I am going to solve for
yy size 12{y} {} by elimination. I will use all the exact same steps we have always used in class.
To do this, I will multiply the top equation by
cc size 12{c} {} and the bottom equation by
aa size 12{a} {}.
acx
+
bcy
=
ec
acx
+
bcy
=
ec
size 12{ ital "acx"+ ital "bcy"= ital "ec"} {}
acx
+
ady
=
af
acx
+
ady
=
af
size 12{ ital "acx"+ ital "ady"= ital "af"} {}
This will make the
xx size 12{x} {} terms go away.
acx
+
bcy
=
ec
−
(
acx
+
ady
=
af
)
̲
bcy
−
ady
=
ec
−
af
acx
+
bcy
=
ec
−
(
acx
+
ady
=
af
)
̲
bcy
−
ady
=
ec
−
af
alignl { stack {
size 12{" " ital "acx"+ ital "bcy"= ital "ec"} {} #
size 12{ {underline { - \( "acx"+ ital "ady"= ital "af" \) }} } {} #
size 12{" " ital "bcy" - ital "ady"= ital "ec" - ital "af"} {}
} } {}
This is something we’ve done many times in class, right? First, pull out a
yy size 12{y} {}; then divide by what is in parentheses.
bcy
−
ady
=
ec
−
af
bcy
−
ady
=
ec
−
af
size 12{ ital "bcy" - ital "ady"= ital "ec" - ital "af"} {}
y
=
(
bc
−
ad
)
=
ec
−
af
y
=
(
bc
−
ad
)
=
ec
−
af
size 12{y= \( ital "bc" - ital "ad" \) = ital "ec" - ital "af"} {}
y
=
ec
−
af
bc
−
ad
y
=
ec
−
af
bc
−
ad
size 12{y= { { ital "ec" - ital "af"} over { ital "bc" - ital "ad"} } } {}
We have come up with a totally generic formula for finding
yy size 12{y} {} in any simultaneous equations. For instance, suppose we have…
3x
+
4y
=
18
3x
+
4y
=
18
size 12{3x+4y="18"} {}
5x
+
2y
=
16
5x
+
2y
=
16
size 12{5x+2y="16"} {}
We now have a new way of solving this equation: just plug into
y=ec−afbc−ady=ec−afbc−ad size 12{y= { { ital "ec" - ital "af"} over { ital "bc" - ital "ad"} } } {}. That will tell us that
y=(18)(5)−(3)(16)(4)(5)−(3)(2)=90−4820−6=4214=3y=(18)(5)−(3)(16)(4)(5)−(3)(2)=90−4820−6=4214=3 size 12{y= { { \( "18" \) \( 5 \) - \( 3 \) \( "16" \) } over { \( 4 \) \( 5 \) - \( 3 \) \( 2 \) } } = { {"90" - "48"} over {"20" - 6} } = { {"42"} over {"14"} } =3} {}
Here’s how to find out.
- Do the whole thing again, starting with the generic simultaneous equations, except solve for
xx size 12{x} {} instead of
yy size 12{y} {}.
- Use your formula to find
xx size 12{x} {} in the two equations I did at the bottom (under “So what did we do?”)
- Test your answer by plugging your
xx size 12{x} {} and my
y=3y=3 size 12{y=3} {} into those equations to see if they work!
"This is the "main" book in Kenny Felder's "Advanced Algebra II" series. This text was created with a focus on 'doing' and 'understanding' algebra concepts rather than simply hearing about them in […]"