a question on systems of equations: 4x+y=5 and 7x+3y=10

[allegansveritatem]

Today I was working on a problem. There was a system of 2 equations that had to be solved using substitution. I had done many of these and this system looked pretty simple.

4x+y=5
7x+3y=10

I took y=5-4x and did this:

4x+5-4x=5

and got the true statement:

5=5.

Not being satisfied with this I plugged 5-4x for y into the second equation and got:

7x+3(5-4x)=10
5x+15=10
x=1

To make a long story short I found that the solution to this system was (1,1) when I put the substitute derived from the first equation into the other equation. But when I tried putting the substitute back into the equation it had been developed from....things went strange.

Of course I went back and studied the chapter again and found that you have to use both equations to get this to work. One equation provides the substitute for the other. How funny I missed that and only now after having done this operation 50 times do I learn it. I guess I was just going by rote and luck before.

But, I would like to ask this: Why doesn't the substitute work on both equations? Seems like it should. I mean, y is y whether it is in one equaton or the other, no?

 

[Dr.Peterson]

Today I was working on a problem. There was a system of 2 equations that had to be solved using substitution. I had done many of these and this system looked pretty simple.

4x+y=5
7x+3y=10

I took y=5-4x and did this:

4x+5-4x=5

and got the true statement:

5=5.

Not being satisfied with this I plugged 5-4x for y into the second equation and got:

7x+3(5-4x)=10
5x+15=10
x=1

To make a long story short I found that the solution to this system was (1,1) when I put the substitute derived from the first equation into the other equation. But when I tried putting the substitute back into the equation it had been developed from....things went strange.

Of course I went back and studied the chapter again and found that you have to use both equations to get this to work. One equation provides the substitute for the other. How funny I missed that and only now after having done this operation 50 times do I learn it. I guess I was just going by rote and luck before.

But, I would like to ask this: Why doesn't the substitute work on both equations? Seems like it should. I mean, y is y whether it is in one equaton or the other, no?

When you substitute the solution of one equation back into the same equation, (a) you are not using the second equation, so you can't possibly be solving the system; and (b) all you are doing is "checking" what you did with the first: if you substituted into the same equation and didn't get a true equation like 1=1, that would mean you made a mistake.

To put it another way, when you substitute back into the first equation, the result says that any value of x will work - in the "system" that consists only of the one equation! When you solve correctly, you find the intersection of the two lines; when you do it wrong, you are finding the intersections of the one line with itself, which means the entire line.

When I teach this method, I label the equations A and B, and emphasize that if you solve A, you plug into B, and if you solve B, you plug into A. It's important to keep track of this, as you found out!

 

[JeffM]

When you rearrange the first equation to get an equivalent equation and then substitute back into the first equation, you have not introduced any new information into the first equation.

When you rearrange the first equation to get an equivalent equation and then use that information in the second equation, you have introduced new information into the second equation.

A different way to say much the same thing is this.

\(\displaystyle 4x + y = 5 \iff y = 5 - 4x.\)

There are an infinite number of pairs of real numbers for which those equations are true, not just (1, 1). For example, both equations are true of (0.8, 1.8).

\(\displaystyle 7x + 3y = 10.\)

Again there are an infinite number of pairs of real numbers for which that equation is true, for example (2/3, 16/9)

But when you substitute (5 - 4x) for y in the second equation, you are putting an extra constraint on the second equation.

Because the equations are linear, consistent, and independent, there is only one pair of numbers that satisfies the second equation while satisfying the extra constraint forced by the substitution

 

[allegansveritatem]

When you substitute the solution of one equation back into the same equation, (a) you are not using the second equation, so you can't possibly be solving the system; and (b) all you are doing is "checking" what you did with the first: if you substituted into the same equation and didn't get a true equation like 1=1, that would mean you made a mistake.

To put it another way, when you substitute back into the first equation, the result says that any value of x will work - in the "system" that consists only of the one equation! When you solve correctly, you find the intersection of the two lines; when you do it wrong, you are finding the intersections of the one line with itself, which means the entire line.

When I teach this method, I label the equations A and B, and emphasize that if you solve A, you plug into B, and if you solve B, you plug into A. It's important to keep track of this, as you found out!

I think part of my problem here is that my author does not stress this point. He merely supplies a list of steps, on e of which is to put the substitute into the other equation. I think he should have made it very clear that this procedure won't work UNLESS both equations are let into the act. I am surprised he didn't make this point more strongly.

 

[allegansveritatem]

When you rearrange the first equation to get an equivalent equation and then substitute back into the first equation, you have not introduced any new information into the first equation.

When you rearrange the first equation to get an equivalent equation and then use that information in the second equation, you have introduced new information into the second equation.

A different way to say much the same thing is this.

\(\displaystyle 4x + y = 5 \iff y = 5 - 4x.\)

There are an infinite number of pairs of real numbers for which those equations are true, not just (1, 1). For example, both equations are true of (0.8, 1.8).

\(\displaystyle 7x + 3y = 10.\)

Again there are an infinite number of pairs of real numbers for which that equation is true, for example (2/3, 16/9)

But when you substitute (5 - 4x) for y in the second equation, you are putting an extra constraint on the second equation.

Because the equations are linear, consistent, and independent, there is only one pair of numbers that satisfies the second equation while satisfying the extra constraint forced by the substitution

Yes. I got a wonky result because I tried to cross one equation with itself...inbreeding!

 

[Bob Brown MSEE]

Just for fun

Yes. I got a wonky result because I tried to cross one equation with itself...inbreeding!

You can literally "cross" multiply one line with the other!

Let the two equations
A: 4x + y - 5 = 0
B: 7x + 3y - 10 = 0

Be represented by two vectors in this way (basis vectors i,j,k)
A = 4i + j - 5k
B = 7i + 3j - 10k

Calculate A x B, (A cross B)
A x B = (28*0 + 12k + 40j) + (-7k + 3*0 - 10i) + (-35j +15i + 50*0)
A x B = 5i + 5j + 5k

Where on the z=1 plane is the projected point?
(5x:5y:5z)=(1x:1y:1z)

so x=1 and y=1