Download Elimination Proof Method - Student AND Teacher

ax + by = c
dx + ey = f
Solve the system of equations using the substitution method justifying each step or by filling in the step
indicated by the justification.
ax + by = c

justification: _____________________

dx + ey = f
𝑦=

by = c – ax
justification: _____________________
𝑏
____________________________

__________________
𝑐−𝑎𝑥
___________________
____________________________
In the previous two steps both equations have been solved for y. Substitute the y-value for one equation in
for the y-value in the second equation. You should now have an equation in terms of x with no y.
𝑐−𝑎𝑥
𝑏
=
𝑓−𝑑𝑥
𝑒
Substitution Property
e (c – ax) = _________________
Multiply both sides by b & e by the Multiplication Property
of Equality – NOTE that this is often referred to as crossmultiplication; however, cross-multiplication is not one of
our properties of equality.
______________ = ______________
Distributive Property completed on both sides as needed
bdx + ec – eax = _______________
Add bdx to both sides by the _________________________
bdx – eax = _______________
____________________ from both sides by the Subtraction
Property of Equality
x (bd – ea) = bf – ec
___________________________________________
x = ___________________
Divide both sides by ___________________by the Division
Property of Equality
Now, you can take either of the original equations OR either of the original equations solved for y to find
the y value of the solution by substituting in the value you found for x.
ax + by = c
dx + ey = f
Solve the system of equations using the elimination method justifying each step or by filling in the step
indicated by the justification. We should get the same solutions as the previous where we solved using the
substitution method. There are infinitely many ways that we could solve this system using the elimination
method so I have chosen one for proof purposes. I would encourage you to try other options!
d (ax + by) = dc
Multiply both sides of the first equation by d by the
________________________________________
__________________
Distributive Property
__________________
Multiply both sides of the second equation by -a by the
________________________________________
__________________
Distributive Property
We will now add the two new equations together. Why are we allowed to do this? This step requires us
to use the Addition Property of Equality AND the Substitution Property simultaneously as we add the lefthand sides of the equations together and the right-hand sides of the equations together. We are using the
Addition Property of Equality because we are actually adding the same thing to both sides. It does NOT
look like we are adding the same thing to both sides, but that is where the Substitution Property comes in!!
adx + bdy = dc
+ -adx + -aey = -af
 bdy – aey =dc – af
_______________________
Factor out the y left-hand side
_______________________
Divide both sides by ______________ by the Division
Property of Equality
y = ___________________
Solution should match y solution from previous method
We can now take the y solution and substitute it in either of the original equations. You should have an
equation with an x value but no y values. It should also match the solution found by substitution. We
have now proved that the elimination method yields the same solutions as the substitution method.
ax + by = c
dx + ey = f
Solve the system of equations while replacing one equation by the sum of that equation and a
multiple of the other showing that it produces a system with the same solutions.
We are going to replace the second equation, dx + ey = f, with the sum of it and a multiple, n, of the first
equation, ax + by = c.
n(ax + by) = nc
Multiply both sides by n by the _______________________
____________
Distributive Property
nax + nby + (dx + ey) = nc + (dx + ey)
Add ________ to both sides by the ____________________
nax + nby + (dx + ey) = nc + f
Substitution Property because (dx + ey) =___
nax + dx + nby + ey = nc + f
Apply Commutative & Associative Properties of _________
(na + d) x + (nb + e) y = nc + f
______________________________
Based on the standard, we are now going to replace our original 2nd equation (dx + ey = f) with the sum of
that one and a multiple of the other that we just found ((na + d) x + (nb + e) y = nc + f) to show that we
get the same solutions as when we solved the system with the original two equations. We can use any
method to do this. We will use substitution.
ax + by = c
(na + d) x + (nb + e) y = nc + f
ax + by = c

by = c – ax

𝑦=
𝑐−𝑎𝑥
𝑏
__________________________
__________________________
(na + d) x + (nb + e) y = nc + f
 (nb + e) y = nc + f – (na + d) x

_______________________
___________________________
_______________________
𝑐−𝑎𝑥
𝑏
= _________________
(nb + e) (c – ax) = b (𝑛𝑐 + 𝑓 − (𝑛𝑎 + 𝑑)𝑥)
Substitution Property
Multiply both sides by b & (nb + e) by the
____________________________________
_____________________________________
Distributive Property completed on both sides
as needed
___________________
Add nbc & nbax to both sides by the
________________________________________
bdx – eax = _________
________________ from both sides by the
Subtraction Property of Equality & Add bdx to both
sides by the Addition Property of Equality
x (bd – ea) = bf – ec
_______________________________
x = ____________
Divide both sides by ___________ by the Division
Property of Equality
WOW! This is the same x that we arrived at earlier using the original equations!! The LAST step is to
find the corresponding y-value. We can substitute the found x-value into the equation ax + by = c.
However, we do not need to do this again! We used this x-value and this equation to find the y-value in
the substitution method at the beginning. You can do it again for practice!!
ax + by = c
dx + ey = f
Solve the system of equations using the substitution method justifying each step or by filling in the step
indicated by the justification.

ax + by = c
by = c – ax

justification: Subtraction Property of Equality

dx + ey = f
ey = f - dx

justification: Subtraction Property of Equality
𝑦=
𝑐−𝑎𝑥
𝑏
Division Property of Equality
𝑦=
𝑓−𝑑𝑥
𝑒
Division Property of Equality
In the previous two steps both equations have been solved for y. Substitute the y-value for one equation in
for the y-value in the second equation. You should now have an equation in terms of x with no y.
𝑐−𝑎𝑥
𝑏
=
𝑓−𝑑𝑥
𝑒
Substitution Property
e (c – ax) = b (f – dx)
Multiply both sides by b & e by the Multiplication Property
of Equality – NOTE that this is often referred to as crossmultiplication; however, cross-multiplication is not one of
our properties of equality.
ec - eax = bf – bdx
Distributive Property completed on both sides as needed
bdx + ec – eax = bf
Add bdx to both sides by the Addition Property of Equality
bdx – eax = bf – ec
Subtract ec from both sides by the Subtraction
Property of Equality
x (bd – ea) = bf – ec
Factor out x from the left
x =
𝑏𝑓−𝑒𝑐
𝑏𝑑−𝑒𝑎
Divide both sides by bd – ea by the Division
Property of Equality
Now, you can take either of the original equations OR either of the original equations solved for y to find
the y value of the solution by substituting in the value you found for x.
**This is an important point to reiterate – the solution is an ordered pair – NOT just the x!
𝑐−𝑎𝑥
𝑦=


𝑏
𝑐−𝑎(
𝑏𝑓−𝑐𝑒
)
𝑏𝑑−𝑎𝑒
Substitution of x-value
𝑏
𝑐
𝑏
𝑐
−
𝑎
(
𝑏𝑓−𝑐𝑒
𝑏 𝑏𝑑−𝑎𝑒
𝑏𝑑−𝑎𝑒
𝑎

(
) −
𝑏 𝑏𝑑−𝑎𝑒

𝑐𝑏𝑑−𝑐𝑎𝑒−𝑎𝑏𝑓+𝑎𝑐𝑒

𝑏(𝑏𝑑−𝑎𝑒)
𝑐𝑏𝑑−𝑎𝑏𝑓
𝑏(𝑏𝑑−𝑎𝑒)
𝑎
)
Property of fractions ( 𝑐 ±
𝑏𝑓−𝑐𝑒
(
)
𝑏 𝑏𝑑−𝑎𝑒
𝑏
𝑐
=
𝑎±𝑏
𝑐
) **Apply this in reverse!
Common denominator
Distributive property – don’t forget to distribute “ – ”
Commutative/Associative properties of multiplication
and like terms sum to zero


𝑏(𝑐𝑑−𝑎𝑓)
𝑏(𝑏𝑑−𝑎𝑒)
𝑐𝑑−𝑎𝑓
𝑏𝑑−𝑎𝑒
=𝑦
factor out b from numerator
b factor in numerator and denominator divide to 1
𝑏𝑓−𝑒𝑐 𝑐𝑑−𝑎𝑓
We found y!! Therefore the solution to our system is (𝑏𝑑−𝑒𝑎 , 𝑏𝑑−𝑎𝑒).
We are now going to solve the same system using the elimination method, which will prove that the
elimination method yields the same solution as the substitution method. This still will not cover the
standard but gets us closer to our goal.
ax + by = c
dx + ey = f
Solve the system of equations using the elimination method justifying each step or by filling in the step
indicated by the justification. We should get the same solutions as the previous where we solved using the
substitution method. There are infinitely many ways that we could solve this system using the elimination
method so I have chosen one for proof purposes. I would encourage you to try other options!
d (ax + by) = dc
Multiply both sides of the first equation by d by the
Multiplication Property of Equality
dax + dby = dc
Distributive Property
-a (dx + ey) = -af
Multiply both sides of the second equation by -a by the
Multiplication Property of Equality
-adx – aey = -af
Distributive Property
We will now add the two new equations together. Why are we allowed to do this? This step requires us
to use the Addition Property of Equality AND the Substitution Property simultaneously as we add the lefthand sides of the equations together and the right-hand sides of the equations together. We are using the
Addition Property of Equality because we are actually adding the same thing to both sides. It does NOT
look like we are adding the same thing to both sides, but that is where the Substitution Property comes in!!
dax + dby = dc
 dax + dby + (-adx – aey) = dc + (-adx – aey)
Addition Property of Equality:
add (-adx – aey) to both sides
 dax + dby + (-adx – aey) = dc + -af
Substitution Property since -adx – aey = -af
Which we often write as follows…
adx + bdy = dc
+ -adx + -aey = -af
 bdy – aey =dc – af
y (bd – ae) = dc – af
𝑦=
𝑑𝑐−𝑎𝑓
𝑏𝑑−𝑎𝑒
Factor out the y left-hand side
Divide both sides by (bd – ae) by the Division
Property of Equality
y=
𝑐𝑑−𝑎𝑓
Solution should match y solution from previous method
𝑏𝑑−𝑎𝑒
Ours matches almost! We just need to apply the Commutative Property of
Multiplication to dc to make it cd.
We can now take the y solution and substitute it in either of the original equations. You should have an
equation with an x value but no y values. It should also match the solution found by substitution. We
have now proved that the elimination method yields the same solutions as the substitution method.
𝑎𝑥 + 𝑏𝑦 = 𝑐
𝑐𝑑−𝑎𝑓
 𝑎𝑥 + 𝑏 (
𝑏𝑑−𝑎𝑒
)=𝑐
Substitute our y-value into the first original equation
 𝑎𝑥(𝑏𝑑 − 𝑎𝑒) + 𝑏(𝑐𝑑 − 𝑎𝑓) = 𝑐(𝑏𝑑 − 𝑎𝑒)
Multiply through by (bd – ae)
Multiplication Property of Equality
 𝑎𝑏𝑑𝑥 − 𝑎2 𝑒𝑥 + 𝑏𝑐𝑑 − 𝑏𝑎𝑓 = 𝑐𝑏𝑑 − 𝑐𝑎𝑒
Distributive Property
 𝑎𝑏𝑑𝑥 − 𝑎2 𝑒𝑥 − 𝑏𝑎𝑓 = −𝑐𝑎𝑒
Subtract bcd (cbd) from both sides by the
Subtraction Property of Equality
 𝑎𝑏𝑑𝑥 − 𝑎2 𝑒𝑥 = 𝑏𝑎𝑓 − 𝑐𝑎𝑒
Add baf to both sides by the Addition Property of
Equality
 𝑎𝑥 (𝑏𝑑 − 𝑎𝑒) = 𝑎(𝑏𝑓 − 𝑐𝑒)
Factor out common factors
 𝑥 (𝑏𝑑 − 𝑎𝑒) = 𝑏𝑓 − 𝑐𝑒
Divide through by a by the Division Property of
Equality
 𝑥=
 𝑥=
𝑏𝑓−𝑐𝑒
Divide through by (bd – ae) by the Division
Property of Equality
𝑏𝑑−𝑎𝑒
𝑏𝑓−𝑒𝑐
Apply the Commutative Property of Multiplication
to ce and ae so that it matches our previous solution
𝑏𝑑−𝑒𝑎
𝑏𝑓−𝑒𝑐 𝑐𝑑−𝑎𝑓
The solution to our system is (𝑏𝑑−𝑒𝑎 , 𝑏𝑑−𝑎𝑒).
ax + by = c
dx + ey = f
Solve the system of equations while replacing one equation by the sum of that equation and a multiple of
the other showing that it produces a system with the same solutions.
We are going to replace the second equation, dx + ey = f, with the sum of it and a multiple, n, of the first
equation, ax + by = c.
n(ax + by) = nc
Multiply both sides by n by the Multiplication Property of
Equality
nax + nby = nc
Distributive Property
nax + nby + (dx + ey) = nc + (dx + ey)
Add (dx + ey) to both sides by the Addition Property of
Equality
nax + nby + (dx + ey) = nc + f
Substitution Property because (dx + ey) = f
nax + dx + nby + ey = nc + f
Apply Commutative & Associative Properties of Addition
(na + d) x + (nb + e) y = nc + f
Factor out common factors of x & y
Based on the standard, we are now going to replace our original 2nd equation (dx + ey = f) with the sum of
that one and a multiple of the other that we just found ((na + d) x + (nb + e) y = nc + f) to show that we
get the same solutions as when we solved the system with the original two equations. We can use any
method to do this. We will use substitution.
ax + by = c
(na + d) x + (nb + e) y = nc + f
ax + by = c

by = c – ax

𝑦=
𝑐−𝑎𝑥
𝑏
Subtraction Property of Equality
Division Property of Equality
(na + d) x + (nb + e) y = nc + f
 (nb + e) y = nc + f – (na + d) x
Subtraction Property of Equality

𝑐−𝑎𝑥
𝑏
=
𝑦=
𝑛𝑐 + 𝑓 − (𝑛𝑎 + 𝑑)𝑥
Division Property of Equality
𝑛𝑏 + 𝑒
𝑛𝑐 + 𝑓 − (𝑛𝑎 + 𝑑)𝑥
Substitution Property
𝑛𝑏 + 𝑒
(nb + e) (c – ax) = b (𝑛𝑐 + 𝑓 − (𝑛𝑎 + 𝑑)𝑥)
Multiply both sides by b & (nb + e) by the
Multiplication Property of Equality
nbc – nbax + ec – aex = nbc + bf – nbax – bdx
Distributive Property completed on both sides
as needed
ec – aex = bf – bdx
Add nbc & nbax to both sides by the Addition
Property of Equality
bdx – eax = bf – ec
Subtract ec from both sides by the Subtraction
Property of Equality & Add bdx to both sides by the
Addition Property of Equality
x (bd – ea) = bf – ec
Factor out x from the left
x =
𝑏𝑓−𝑒𝑐
𝑏𝑑−𝑒𝑎
Divide both sides by bd – ea by the Division
Property of Equality
WOW! This is the same x that we arrived at earlier using the original equations!! The LAST step is to
find the corresponding y-value. We can substitute the found x-value into the equation ax + by = c.
However, we do not need to do this again! We used this x-value and this equation to find the y-value in
the substitution method at the beginning. You can do it again for practice!!