Lesson HOW TO algebreze and solve these problems?

HOW TO algebreze and solve these problems?

Problem 1

Let m = # of pupils who took the mathematics exam, and 
    g = # of pupils who took the geology     exam.

We are given that

m - g = 50,    (1)

it is our first equation. Now,

"A fifth of those who took the mathematics exam were girls" - their number is .   (2)

"A quarter of those who took the geology exam were girls" - their number is .     (3)

We also are given that 

"the number of girls who took the mathematics exam was six more than the number of girls who took the geology exam".

In other words, the number (2) is 6 more than the number (3),  i.e.

 = 6.   (4)

Thus we have now the system of two equations in two unknowns (1) and (4). 
Let me rewrite them one more time:

m - g = 50,  (1)

 = 6.   (4)

Actually, the problem is just solved - the system of equations is established. The rest is just technique.

To solve the system, multiply (1) by 4 and multiply (4) by 20 (both sides). You will get

4m - 4g = 200,   (1')

4m - 5g = 120.   (4')

Next, distract (4') from (1'). You will get

g = 200 - 120 = 80.

Then from (1),  m = 50 + g = 50 + 80 = 130.

Answer. The number of pupils who took the maths exam is 130.

Problem 2

It will be easier if we find out how many boys & girls are in the school.
Let b = no. of boys in the school
Let g = no. of girls
:
b + g = 1025
the number of girls & boys who can't swim are given as 400
b + g = 400


Get rid of the denominators, multiply by 35
7b + 20g = 14000
multiply the 1st equation by 7, subtract from the above equation
7b + 20g = 14000
7b +  7g = 7125
------------------subtraction eliminates b, find g
0 + 13g = 6825
g = 6825/13
g = 525 girls in the school
then
1025 - 525 = 500 boys in the school
"If x boys can swim, write an equation for x and solve it."
x = 500(- )
x = 500()
x = 400 boys can swim