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# simultaneous equations

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how to solve symultanious equations

5x+2y=-2

3x-5y=17.4

yahya  May 13, 2017
edited by yahya  May 13, 2017
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#1
+6365
+3

Can you give an example of one of these problems?

hectictar  May 13, 2017
#2
+6928
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5x + 2y = -2 --- call this the first equation

3x - 5y = 17.4 --- call this the second equation

First equation implies that

3x + 1.2y = -1.2 --- call this the third equation

Subtract the second equation from the third equation and get:

6.2 y = -18.6

y = -3

Substitute the answer for y into the first equation and get:

5x - 6 = -2

5x = 4

x = 4/5

And done!!

MaxWong  May 15, 2017
#3
+6928
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And if you want to solve a general linear simultaneous equation: (That's what we do in 8th grade...... :) )

ax + by = c --- (1)

dx + ey = f --- (2)

(1) implies that:

y = (c - ax)/b --- (3)

Substitute (3) into (2):

dx + (ec - aex)/b = f

dx + ec/b - aex/b = f

dx - aex/b = f - ec/b <---- from now on I have to use LaTeX.... Too complicated to type like that....

$$x(d - \dfrac{ae}{b})=f - \dfrac{ce}{b}\\ x = \dfrac{f - \frac{ce}{b}}{d - \frac{ae}{b}} = \dfrac{bf-ce}{bd-ae}$$

That's the general solution for x. Now let's work that out for y.

(1) implies that

$$x = \dfrac{c-by}{a}$$ --- (4)

Substitute (4) into (2):

$$\dfrac{dc-bdy}{a}+ey = f\\ \dfrac{dc}{a}-y\cdot \dfrac{bd}{a} + ey = f\\ y(e-\dfrac{bd}{a})=f-\dfrac{cd}{a}\\ y = \dfrac{f-\frac{cd}{a}}{e-\frac{bd}{a}}=\dfrac{af-cd}{ae-bd}$$

Next time you see a simultaneous equation, plug those values in, done!!

MaxWong  May 15, 2017

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