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EQUATIONS
SOLVE
1. 2x+9=13
2. (6x+9)/3= 4
3. 5y+4 = 3(y+8)-5
4. 2(c-1)- (4c+6)+8(c+1)= 36
5. (x+8)/(x+4)=2
6. (x+5)/2 + (x+5)/2 = 10
7. 2(t-1)+ 3t = t+14
8. 2(2x+1)= 15-3x
9. 3x+4 = 2(x+2)
10. 2(4-x)+5(2-x)=0
SOLUTION OF EQUATIONS INVOLVING ALGEBRAIC FRACTIONS
ex:-
1/(2x+1) = 1/3(x-5)
The lowest common multiple of the terms (2x+1) and 3(x-5) in the denominators is 3(2x+1)(x-5)
Let us multiply each term of the equation by this lowest common multiple.
1/(2x+1) = 1/3(x-5)
3(x-5) = 2x+1
3x-15 = 2x+1
3x-2x = 1+15
x = 16
Solve the following equations.
1. 15/x + 2/x =17
2. 1/x + 1/3x =1
3. 1/(x-1) + 1/2(x-1) =5
4. x/2 + x/6 =16
5. 1/(x-2) + 1/(x+5) = 0
6. 3/(10+x) = 1/x
7. 2(x+1)/3 + 3(x-5)/5 =1/15
8. (x-7)/2 = 2(x-2)/5
CONSTRUCTION OF SIMPLE EQUATIONS AND THEIR SOLUTIONS.
A father is thrice as old as his son. In another ten years, the father will be twice as old as his son. Find the present
ages of the father and the son.
Let the age of the son be x.
Then the age of the father =3x
In ten years the age of the son = x+10
In ten years the age of father = 3x + 10
In ten years as the father will be twice as old as his son,
2(x+10) = 3x +10
The age of the son and the father can be determined by solving this equation.
therefore 2(x+10) =3x+10
2x+20 =3x+10
20-10 =3x-2x
10 =x
adding -10 and -2x to both sides.
therefore present age of son = 10 years
present age of father = 30 years |
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