Basic Simultaneous Equations Problem

[paulb203]

Homework Statement

2c+3k=5.3
4c+4k=8.8
c=?

Relevant Equations

N/A

I got c=13, but I'm using an app that, at the 'diagnostic' stage, doesn't tell you if you're correct or not
(I multiplied the first equation by 4, and the second by 2, then subtracted the second from the first)

 

[Bosko]

By multiplying the first by 4 and the second by -3 we get
$8c+12k=21.2$
$-12c-12k=-26.4$
and by adding them together you can obtain $c$.

 

[Mark44]

Homework Statement: 2c+3k=5.3
4c+4k=8.8
c=?
Relevant Equations: N/A

I got c=13, but I'm using an app that, at the 'diagnostic' stage, doesn't tell you if you're correct or not
(I multiplied the first equation by 4, and the second by 2, then subtracted the second from the first)

c = 13 is not correct.
You can verify that a solution pair is correct by substituting the values you found into the equations. If the pair is correct you will get the same numbers on both sides of each equation.

 

[Mark44]

By multiplying the first by 4 and the second by -3 we get
$8c+12k=21.2$
$-12c-12k=-26.4$
and by adding them together you can obtain $c$.

Or you can multiply the second equation by -1/2, then add the equations to get one equation in d only.

 

[Steve4Physics]

2c+3k=5.3
4c+4k=8.8

Inspecting the equations you should notice that if you multiply the 1st equation by 2 then you get 4c which is the same as in the second equation. So do that.

Then subtract equations to get an equationfor k.

Solve for k.

Substitute your value of k in one of the equations to get an equation for c.

Solve for c.

Check your values for k and c satisfy the original equations.

You shouldn't need an app!

 

[paulb203]

By multiplying the first by 4 and the second by -3 we get
$8c+12k=21.2$
$-12c-12k=-26.4$
and by adding them together you can obtain $c$.

Thanks, Bosko. Why did you choose those numbers? I tried them out and got the same answer as I got. Nb; there is a typo in my original post; it should read 1.3, not 13. I tried to edit it but found out there is a 360 minutes time limit.

 

[paulb203]

c = 13 is not correct.
You can verify that a solution pair is correct by substituting the values you found into the equations. If the pair is correct you will get the same numbers on both sides of each equation.

Thanks, Mark. Typo. Should read 1.3, not 13.

 

[paulb203]

Inspecting the equations you should notice that if you multiply the 1st equation by 2 then you get 4c which is the same as in the second equation. So do that.

Then subtract equations to get an equationfor k.

Solve for k.

Substitute your value of k in one of the equations to get an equation for c.

Solve for c.

Check your values for k and c satisfy the original equations.

You shouldn't need an app!

Thanks, Steve. That makes sense. I did have some (faulty) logic for choosing to do it that way, but I forget what it was. I tried it your way and got 1.3, which is the same as the answer I got first time around (typo in the original post; should've said 1.3, not 13).

 

[Steve4Physics]

Thanks, Steve. That makes sense. I did have some (faulty) logic for choosing to do it that way, but I forget what it was. I tried it your way and got 1.3, which is the same as the answer I got first time around (typo in the original post; should've said 1.3, not 13).

I just realised that you only want to find c. Not k.

So @Mark44's (Post #4) method is more efficient.

 

[Bosko]

Thanks, Bosko. Why did you choose those numbers?

You need $c$. I want to eliminate $k$ by setting numbers in front of $k$ in both equations to be the same but with opposite signs.
You can use other numbers to achieve the same goal.

2c+3k=5.3
4c+4k=8.8
c=?

By multiplying $3k$ by $4$ I got $12k$ in the first and by multiplying $4k$ with $-3$ I got $-12k$ in the second equation.

By adding both equation we get the single equation without $k$.

Edit: There are many other way to solve those equations

For example, you can notice that the second equation can be easily divided by 4 and we get
$2c+3k=5.3$
$c+k=2.2$

then also, from the second equation we can express $k$ as
$k=2.2-c$

and put this ($2.2-c$) instead of $k$ in the first equation
$2c+3(2.2-c)=5.3$
$2c+6.6-3c=5.3$