ok sove for x in terms of y in #1:Originally Posted by danielwu
then substitute that value in for x in #2
does that help???
dan
Hi guys after the lessons i have problem understanding and solving system of linear equations. I need help badly to understand how it actually work outs???
pls help me
solve the given systems of equations by the method of elimination by substitution
33x + 2y = 34
40y = 9x + 11
by the method of elimination by addition or subtraction
2x - 3y - 4 = 0
3x + 2 = 2y
Because you are going to subtract the two resulting equations and eliminateOriginally Posted by danielwu
the term in y. So you need to make the coefficient of y in each equation the
same.
In the first equation the coefficient of y is -3, and in the second is -2.
So multiplying these equations by 2 and 3 makes the coefficient of y -6
in both equations, so when the equations are subtracted the y terms is
removed.
RonL
hi i am still quite confuse.
it says solve the given systems of equations by either method of this section?
30p = 55 - Q
19p + 14Q + 3 = 0
so in becomes
30p + Q = 55
19p + 14Q = -3
and from equation (1) do i multiply by 14? to make the coefficient of Q in (1) the same as equation 2?
420p + 14Q = 770
That is correct, now subtract the second equation fron this to get:Originally Posted by danielwu
420p - 19p + 14Q - 14Q = 770 + 3,
simplifying:
401p = 773,
p = 773/401.
Now going back to your first equation:
30p + Q = 55,
so:
Q=55-30p=55-30.773/401=-1135/401
RonL