How would I do...
4p-3q=6
3p-2q=5
2x=3y-2
3y=1+4x
and
4c + 3t =515
3c + 5t = 565
Thanks a lot. Still havent got my head around Simultaneous Equations
You have equation: $\displaystyle 3y=1+4x$
Subtract 2x from both sides: $\displaystyle 3y-2x=1+4x-2x$
But look at your first equation, you know that 2x equals 3y-2. So substitute: $\displaystyle 3y-(3y-2)=1+4x-2x$
Simplify: $\displaystyle 3y-3y+2=1+2x$
Therefore: $\displaystyle 2=1+2x$
Subtract 1 from both sides: $\displaystyle 1=2x$
divide by 2: $\displaystyle \frac{1}{2}=x$
Go back to original equation: $\displaystyle 2x=3y-2$
Substitute: $\displaystyle 2\left(\frac{1}{2}\right)=3y-2$
Multiply: $\displaystyle 1=3y-2$
Add 2 to both sides: $\displaystyle 3=3y$
Divide by 3: $\displaystyle 1=y$
Can you do the other ones?
You want to modify the equations so that one of the variables has the same
coefficient in both the equations, then by subtracting the equations you
can eliminate that variable.
In this case if we multiply the first equation by 3 and the second by 4 the
coefficient of p will be 12 in both cases:
12p - 9q = 18
12p - 8q = 20.
Subtracting these equations gives:
(12p - 9q) - (12p - 8q) = 18-20,
or:
-q = -2,
so q=2.
Now we substitute this value of q back into any of the equations that
we have to find the value of p. So lets take the equation
4p - 3q = 6,
with q=2 this becomes:
4p - 6 = 6,
or:
4p = 12,
so p=3.
If you reorganise your equations into the standard form with all the unknowns on one side
(and in the same order in all the equations) and the constants on the other side like the
above equations, then this method will work for all your problems.
It has another advantage, in that it can be used with a few more tricks for equations in
more than two variables (of course with the same number of equations as variables).
RonL