Someone please explain in the most simple way possible please. (edit: i am using elimination)
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In: 1
When you write an equation like x + 7 = 9, you’re saying “there’s some unknown number, which I’m going to represent with an x. I don’t know what it is yet, but I know if I add 7 to it, I get 9”. And solving the equation is trying to find values of x that would make that statement true. x=2 is a solution because 2 + 7 = 9 is a true equation.
Simultaneous equations are the same idea, except that you have *multiple* pieces of information about your variables and (typically) more than one variable. So if, for example, you have:
* -2x + 3y = 7
* 3y = 9
then what I’m saying is:
* There are two numbers I don’t know. I’m writing one of them as x and the other as y.
* I know that if I take minus two times x, and add 3 times y, I get 7.
* I know that if I take three times y, I get 9.
* I want to know what x and y are.
——
There is a general fact in math that is very helpful in solving equations:
* If you take a true equation, and do the same thing to both sides, you still have a true equation. In particular, any value of the variables that solved the first equation will solve the new one, too.
And in particular:
* If you take a true equation, and do the same **reversible** thing to both sides, not only do you still have a true equation, but the solutions to the new equation are the same as the solutions to the old equation.
Reversible operations include things like adding, subtracting, or multiplying or dividing by anything but zero. And in this case, those are the operations we’re going to use.
—–
So. We have our equations, and our goal is to transform them into equations we know how to solve. There are a few ways to go about this, but you specified elimination, so we’ll use that.
Our goal is to take the equations involving *two* variables and turn them into an equation involving *one* variable, because we know how to solve one-variable equations.
Let’s consider the first equation, -2x + 3y = 7. Wouldn’t it be nice if we could just get rid of that 3y somehow, so we just had something with *x*? We can subtract 3y from both sides if we want, but that doesn’t help us, because the 3y just ends up on the other side.
Instead, we need to figure out how to cancel out the *y* on the left without introducing a *y* on the right. At first, this might seem impossible if we’re doing the same thing on both sides. But the trick is to remember that just because we’re doing the same thing to both sides doesn’t mean we have to **write** it the same way.
What if we took our first equation, -2x + 3y = 7, and subtracted 3y from the left side and 9 from the right side? We wouldn’t normally be allowed to do this, but because our other equation tells us that 3y = 9, *in this case* we’d actually be doing the same thing to both sides, just written differently.
Then we’d get -2x + 3y – 3y = 7 – 9. Now the 3y’s on the left side cancel out, and we’re left with -2x = -2, which is an equation we know how to solve.
For shorthand, we often say we “subtracted the second equation from the first”, and write it as:
-2x + 3y = 7
-( 3y = 9)
-2x = -2
but what we’re really doing is just subtracting the same value, written in two different ways, from both sides of the first equation.
Following so far?
When you write an equation like x + 7 = 9, you’re saying “there’s some unknown number, which I’m going to represent with an x. I don’t know what it is yet, but I know if I add 7 to it, I get 9”. And solving the equation is trying to find values of x that would make that statement true. x=2 is a solution because 2 + 7 = 9 is a true equation.
Simultaneous equations are the same idea, except that you have *multiple* pieces of information about your variables and (typically) more than one variable. So if, for example, you have:
* -2x + 3y = 7
* 3y = 9
then what I’m saying is:
* There are two numbers I don’t know. I’m writing one of them as x and the other as y.
* I know that if I take minus two times x, and add 3 times y, I get 7.
* I know that if I take three times y, I get 9.
* I want to know what x and y are.
——
There is a general fact in math that is very helpful in solving equations:
* If you take a true equation, and do the same thing to both sides, you still have a true equation. In particular, any value of the variables that solved the first equation will solve the new one, too.
And in particular:
* If you take a true equation, and do the same **reversible** thing to both sides, not only do you still have a true equation, but the solutions to the new equation are the same as the solutions to the old equation.
Reversible operations include things like adding, subtracting, or multiplying or dividing by anything but zero. And in this case, those are the operations we’re going to use.
—–
So. We have our equations, and our goal is to transform them into equations we know how to solve. There are a few ways to go about this, but you specified elimination, so we’ll use that.
Our goal is to take the equations involving *two* variables and turn them into an equation involving *one* variable, because we know how to solve one-variable equations.
Let’s consider the first equation, -2x + 3y = 7. Wouldn’t it be nice if we could just get rid of that 3y somehow, so we just had something with *x*? We can subtract 3y from both sides if we want, but that doesn’t help us, because the 3y just ends up on the other side.
Instead, we need to figure out how to cancel out the *y* on the left without introducing a *y* on the right. At first, this might seem impossible if we’re doing the same thing on both sides. But the trick is to remember that just because we’re doing the same thing to both sides doesn’t mean we have to **write** it the same way.
What if we took our first equation, -2x + 3y = 7, and subtracted 3y from the left side and 9 from the right side? We wouldn’t normally be allowed to do this, but because our other equation tells us that 3y = 9, *in this case* we’d actually be doing the same thing to both sides, just written differently.
Then we’d get -2x + 3y – 3y = 7 – 9. Now the 3y’s on the left side cancel out, and we’re left with -2x = -2, which is an equation we know how to solve.
For shorthand, we often say we “subtracted the second equation from the first”, and write it as:
-2x + 3y = 7
-( 3y = 9)
-2x = -2
but what we’re really doing is just subtracting the same value, written in two different ways, from both sides of the first equation.
Following so far?
Let’s say you’ve got equation (1): 2y = 3x + 7 and equation (2): 5y -3x = 10. You can’t solve an equation with two variables in it so you need to eliminate one of the variables. One way to do this is to add 3x to equation 2 giving you 5y = 3x +10. Now if you do (2) – (1) you’ll get (2y – 5y) = (3x – 3x) + (7 -10) or -3y = -17. We’ve eliminated the x variable so now we’ve got a simple equation with one variable to solve and get y =17/3. Now that we’ve got a value for one variable we can sub the sub value into either equation to get another equation with one variable. Subbing into (1) gives 2 ×17/3 = 3x +7 which can be solved to give x = 9.
Hope this helps
You know the equation of a line, right?
Well, what if you had 2 lines and you wanted to know the x & y coordinate where they meet (if at all)? You could graph it, but you can’t tell if they cross at x=5 or x=5.02 (unless using a computer).
Substitution & elimination are ways to solve it algebraically.
[**I wrote a how-to**](https://docs.google.com/document/d/1OevWqMCtjjAhEy20W1YTwmyyyQIW0WzZ6LvhcrGGdK0/export?format=pdf).
____
____
To use the example from my write-up:
4x – 3y = 5
3x + 2y = 8
____
**Substitution**
If we isolate x in the top equation we get:
x = (5 + 3y)/4
Since their x & y values have to be identical, we can plug (substitute) this “value” of x into the second equation, leaving only y variables which we can isolate:
3( (5 + 3y)/4) ) + 2y = 8
(15 + 9y)/4 + 2y = 8
(15 + 9y)/4 = 8 – 2y
15 + 9y = 32 – 8y
17y = 17
**y = 1**
We can now plug this into either 2 equations (including the isolated x equation):
x = (5 + 3y)/4
x = (5 + 3)/4
x = 8/4
**x = 2**
Thus **(2,1)** is where they cross.
You could have isolated y first, it doesn’t matter.
_____
**Elimination**
Just like substitution, we want to get rid of 1 variable so we can isolate the other. However, this way is different in that you combine the equations, turning them into equation equations if need be.
4x – 3y = 5
3x + 2y = 8
We can multiple the coefficients of a selected variable to obtain a GCF/LCM, in this case multiple the 1st & 2nd equations by 3 & 4 respectively for x :
12x – 9y = 15
12x + 8y = 32
We can now subtract the equations:
-17y = -17
**y = 1**
We can now substitute that in or just re-do the elimination for y, I’ll do the latter by multiplying by 8 & 9 respectively:
96x – 72y = 120
108x + 72y = 288
We can now add the equations
204x = 408
**x = 2**
Thus **(2,1)** is where they cross.
____
When you have 3+ equations and/or 3+ variables, it’s recommend to do Gauss-Jordan elimination visualization, as it saves a lot of space (you just write the coefficients & constants, not the variables, sort of like synthetic division). I can’t give you an example right now as it’s a pain to type. However, here is TOCT’s video on it: https://youtu.be/eYSASx8_nyg
Let’s say you’ve got equation (1): 2y = 3x + 7 and equation (2): 5y -3x = 10. You can’t solve an equation with two variables in it so you need to eliminate one of the variables. One way to do this is to add 3x to equation 2 giving you 5y = 3x +10. Now if you do (2) – (1) you’ll get (2y – 5y) = (3x – 3x) + (7 -10) or -3y = -17. We’ve eliminated the x variable so now we’ve got a simple equation with one variable to solve and get y =17/3. Now that we’ve got a value for one variable we can sub the sub value into either equation to get another equation with one variable. Subbing into (1) gives 2 ×17/3 = 3x +7 which can be solved to give x = 9.
Hope this helps
You know the equation of a line, right?
Well, what if you had 2 lines and you wanted to know the x & y coordinate where they meet (if at all)? You could graph it, but you can’t tell if they cross at x=5 or x=5.02 (unless using a computer).
Substitution & elimination are ways to solve it algebraically.
[**I wrote a how-to**](https://docs.google.com/document/d/1OevWqMCtjjAhEy20W1YTwmyyyQIW0WzZ6LvhcrGGdK0/export?format=pdf).
____
____
To use the example from my write-up:
4x – 3y = 5
3x + 2y = 8
____
**Substitution**
If we isolate x in the top equation we get:
x = (5 + 3y)/4
Since their x & y values have to be identical, we can plug (substitute) this “value” of x into the second equation, leaving only y variables which we can isolate:
3( (5 + 3y)/4) ) + 2y = 8
(15 + 9y)/4 + 2y = 8
(15 + 9y)/4 = 8 – 2y
15 + 9y = 32 – 8y
17y = 17
**y = 1**
We can now plug this into either 2 equations (including the isolated x equation):
x = (5 + 3y)/4
x = (5 + 3)/4
x = 8/4
**x = 2**
Thus **(2,1)** is where they cross.
You could have isolated y first, it doesn’t matter.
_____
**Elimination**
Just like substitution, we want to get rid of 1 variable so we can isolate the other. However, this way is different in that you combine the equations, turning them into equation equations if need be.
4x – 3y = 5
3x + 2y = 8
We can multiple the coefficients of a selected variable to obtain a GCF/LCM, in this case multiple the 1st & 2nd equations by 3 & 4 respectively for x :
12x – 9y = 15
12x + 8y = 32
We can now subtract the equations:
-17y = -17
**y = 1**
We can now substitute that in or just re-do the elimination for y, I’ll do the latter by multiplying by 8 & 9 respectively:
96x – 72y = 120
108x + 72y = 288
We can now add the equations
204x = 408
**x = 2**
Thus **(2,1)** is where they cross.
____
When you have 3+ equations and/or 3+ variables, it’s recommend to do Gauss-Jordan elimination visualization, as it saves a lot of space (you just write the coefficients & constants, not the variables, sort of like synthetic division). I can’t give you an example right now as it’s a pain to type. However, here is TOCT’s video on it: https://youtu.be/eYSASx8_nyg
Two or more equations.
Same variables.
Same values make each equation work.
If you were to draw the graph of both functions you would find one or more points where they meet (intersect). These values of the variables are the solutions to the equations.
You can also determine them algebraically by a variety of methods.
e.g. you have two equations: y = x + 2 and 2x + y = 8
You can look at the first one and say “well, since y = x + 2 I can replace (substitute) the y in the second equation with x + 2” – i.e. 2x + x + 2 = 8 or 3x + 2 = 8
The purpose of doing this is that you now have an equation with only one unknown (x) rather than two (x and y) so it’s very easy to solve: Subtract 2 from each side to leave the ‘x’ term on its own (3x = 6) and then divide both sides by the coefficient (number in front of the x). Dividing both sides by 3 gives x = 2
Now you know x you can look at the other unknown, y. Again y = x + 2; plug in the 2 that you discovered was the value of x: y = 2 + 2 = 4
So x = 2 and y = 4 are the solutions to that pair of simultaneous equations – and if you drew the two on the same axes the lines would intersect at the point (2,4).
Two or more equations.
Same variables.
Same values make each equation work.
If you were to draw the graph of both functions you would find one or more points where they meet (intersect). These values of the variables are the solutions to the equations.
You can also determine them algebraically by a variety of methods.
e.g. you have two equations: y = x + 2 and 2x + y = 8
You can look at the first one and say “well, since y = x + 2 I can replace (substitute) the y in the second equation with x + 2” – i.e. 2x + x + 2 = 8 or 3x + 2 = 8
The purpose of doing this is that you now have an equation with only one unknown (x) rather than two (x and y) so it’s very easy to solve: Subtract 2 from each side to leave the ‘x’ term on its own (3x = 6) and then divide both sides by the coefficient (number in front of the x). Dividing both sides by 3 gives x = 2
Now you know x you can look at the other unknown, y. Again y = x + 2; plug in the 2 that you discovered was the value of x: y = 2 + 2 = 4
So x = 2 and y = 4 are the solutions to that pair of simultaneous equations – and if you drew the two on the same axes the lines would intersect at the point (2,4).
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