# Math Help - solve for x and y

1. ## solve for x and y

solve for x and y

a) (x/3) + (y/4) = 11

(5x/6) - (y/3) + 7 =0

b) 6x + 5y = 7x + 3y + 1 = 2(x + 6y -1)

c) [( x + y - 8 )/2] = [ (x + 2y -14) / 3] = [ ( 3x + y - 12) /11]

d) [(bx)/a] - [(ay)/b] + a + b = 0

bx - ay + 2ab = 0

now add all the values of x an y (i can do this part by myself)

2. Originally Posted by saha.subham
solve for x and y

a) (x/3) + (y/4) = 11

(5x/6) - (y/3) + 7 =0

b) 6x + 5y = 7x + 3y + 1 = 2(x + 6y -1)

c) [( x + y - 8 )/2] = [ (x + 2y -14) / 3] = [ ( 3x + y - 12) /11]

d) [(bx)/a] - [(ay)/b] + a + b = 0

bx - ay + 2ab = 0

now add all the values of x an y (i can do this part by myself)

which method are you required to do them in?

a) make the equations, so they don't have fractions.
for example, first line (x/3) + (y/4) = 11
times it by 3 and 4, so we have 4x+3y = 132

b) 6x + 5y = 2(x + 6y -1)
7x + 3y + 1 = 2(x + 6y -1)

2(x + 6y -1)=2x+12y-2

put all the x and y on one side.
for example, the first line would be 4x-7y=-2

3. Can you solve these for x and y?
2x + y = 7
x + 3y = 11

If not, there's no point in giving you answers for what you posted.

4. Originally Posted by Wilmer
Can you solve these for x and y?
2x + y = 7
x + 3y = 11

If not, there's no point in giving you answers for what you posted.
y= -3
x= +5

5. ## solve for x and y

Originally Posted by Cezzar
y= -3
x= +5
Hello Cezzer,
Have you studied the solution of simultaneous equations? You must eliminate x or y from the two and then use that value in either equation to find the other value. Your answer satisfies only one equation so is wrong

bjh

6. sorry but still i have not got a perfect answer

i am very new in this chapter so i am a layman in this so plzz help me

7. i have solved part b and c thanks to baby millo

now if i could get full solve of part a and d it would be very help full

8. Originally Posted by saha.subham
i have solved part b and c thanks to baby millo

now if i could get full solve of part a and d it would be very help full
can you say what dont you understand in part a?

(x/3) + (y/4) = 11
times it by 3 and 4, so we have $4x+3y = 132$

$\frac{x}{3} + \frac{y}{4} = 11$
as i've said times it by 3.
$3*\frac{x}{3} + 3* \frac{y}{4} = 3*11$
$= x + \frac{3y}{4} = 33$
then times it by 4.
$4*x+ 4*\frac{3y}{4} = 4*33$
$= 4x+3y = 132$

do this to $\frac{5x}{6} -\frac{y}{3} = -7$
so this time times it by 6 and 3.

9. Originally Posted by saha.subham
i have solved part b and c thanks to baby millo

now if i could get full solve of part a and d it would be very help full
d) as im not very good with Algebra, someone else might be better off. ie there is a simpler way.

but here is a method, the answer should come to x=-a , y=b

$\frac{bx}{a} - \frac{ay}{b}+a+b=0$

times it by a and b

$b^2x-a^2y+a^2b+b^2a=0$[1]

the other equation is $bx-ay+2ab=0$ [2]

times the equation by b or a, in my case b.

$b^2x-aby+2ab^2=0$ [2]

then equation $[2]-[1]$

$= (ab-a^2)y= 2ab^2-(a^2b+b^2a+2ab^2)$
$
\rightarrow y= \frac{ 2ab^2-(a^2b+b^2a+2ab^2)}{(ab-a^2)}
$

which can be simplify to $y=b$

sub y in to [2] to find x.

$bx-ay+2ab=0$

$x= \frac{-2ab+ay}{b}$

$given y=b$

$x= \frac{-2ab+ab}{b}$

$x= -2a+a$

$x=-a$

10. ## solve for x and y

originaly posted by saha,subham,

Hello saha,
I hope the following explains fully how similtaneous equations are solved.

equ 1 x/3+y/4 =11 becomes 4x+3y = 132
equ 2 5x/6 -y/3 +7 =0 " 5x -2y +42 =0
in order to solve these two you must multiply each by numbers which will eliminate one variable . in this case use 2 for equ1 and3 for equ 2

1 8x +6y =132
2 15x-6y+126 =0 Add the two equs to get 23x==138 Remember equals added to equals result in equals

x = 6 and y =36 when you plug in the 6 in equ 1 to find y.

I really would like to know if this is clear and helpful to you for handling similar problems

bjh

11. Also Saha, what BJ showed you is what should have been taught
to you in math class before being given these problems; are you
learning on your own? Easier to help if we know...

12. this chapter has not been started yet in our school, so doing by myself now also getting help from my friends.

actually different sums has different problems though same type of sums

so thanks to everybody too many sums are there so more help will be required.