# Kunal math linear equation with one variable

19. May 2014
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### Kunal math linear equation with one variable

• 4. Linear - linear come from the Latin word ‘linearis’ which means created by lines. it consists of one variable and no exponent. For ex - y=25x+100 Equation – A statement in which states two or more algebraic expression are equal. What is a linear equation in one variable The equation involving only one variable in first order is called linear equation. For ex – 2x-3=9
• 5. we know that an algebraic expression is an equation involving Variables.It has an equality sign.the expression on the left of the Equality sign is the left hand side (LHS).the expression on the Right of the equality sign is the right hand side (RHS).
• 6. In an equation the values of the expression on the LHS and RHS equal. This happens to true only for certain values of the Variables these values are the solution of the equation. How to find an solution of an equation we the two sides of the equation are balanced. We perform the same mathematical operation on both Sides of equation ,so that the balance is not disturbed. A few such steps give the solution.
• 7. 1.To solve an equation of the form x+a=b E.x.: Solve x+4=10 Solution: x+4=10 => x+4-4=10-4 (subtracting 4 from both the sides) => x=6 2.To solve an equation of the form x-a=b E.g.: Solve y-6=5 equal. Solution: y-6=5 => y-6+6=5+6 (adding 6 to both sides) => y=11 1. 3)example 1: 2x-3=7 Solution: Adding 3 to both sides. 2x-3+3=7+3 (the balance is not disturbed) 2x=10 Next divide both sides by 2: 2x/2=10/2 (required solution) X=5
• 8. a. In an equation, an added term is transposed (taken) from one side to the other, it is subtracted. i.e.,2y+9=4 Solution: transposing 9 to RHS 2y =4 – 9 2y = -5 (solution) Dividing both sides by 2 , y =-5/2 To check the answer :LHS=2(-5/2)+9=-5+9=4=RHS (as required) b. In an equation, a term in multiplication is transposed to the other side, it is divided. i.e., 3x=12 => x=12/3=4 (3 is transposed). c.In an equation a term in division is taken to the other side it is multiplied. i.e., y/4=6 => y=6×4=24 (4 is transposed) GREAT JOB!!!GREAT JOB!!!
• 9. A. sum of two number is 74 .one number is 10 more than the other .what are the number ans.let first number be x 2nd number be x+10 ATQ: x+(x+10)=74 2x + 10 = 74 2x = 74-10 X= 32 this one is the number. The other number is x+10=32+10=42
• 10. b) Banish has three as many two rupee coins as he has five rupee coin he has an all a sum of RS 77 how many coins of each denomination does he have? Ans –let the number of 5 rupee coins that banish has be x. then the number of two rupee coin he has is three times x or 3x. The amount banish has: From 5 rupee coins,Rs 5*x=5* From 2 rupee coins,Rs 2*3x=6x Hence the total money ha has =11x ATQ: 11X = 77 X = 77/11=7 Thus the number of 5 rupee coin =x=7 Number of two rupee coin =3x=3x7=21
• 11. C .The difference between two number is 66,the ratio of the two number is 5:2 what are the two numbers. Ans. Since the ratio of two number is 2:5 we May take one number be 2x and the other be 5x. ATQ:5x-2=66 3x = 66 x=66/3 x=22 so the number are 5x =2*22 or44 and 5*22 or 110 the difference between them = 110-44=66 as desired
• 12. We know that in the equation 2x-3=7,the two Expression is 2x-3and7. In most examples we come to know that RHS is just a number. But this is not needed always be so; both sides could have expression with variable. For example, the equation 2x-3=x+2 has expression with variable on both sides; the expression on the LHS ism (2x-3) and the expression on the RHS is(x+2).
• 13. a) Solve 2x-3=x+2 Ans or 2x=x+2+3 or 2x=x+5 or 2x-x=x+5-x Here we subtracted from both side of equation not a number constant, but a term involving the variable. We can do this as variable are also number.also,note that subtracting x from both sides amount to transposing x to LHS.
• 14. (b)5t – 3= 3t – 5 Or 5t – 3t= -5 +3 Or 2t=-2 Or t=-2/2 Or t =-1 (c)5x+9=5+3x Or 5x-3x=5-9 Or 2x = -4 Or x = -4/2 Or x = -2
• 15. (a) arjun is twice as old as shiv .five years ago his age was three times shiv age .find their present ages. Ans :let us take shiv present age to be x years Then arjun present age would be 2x years Shiv age five years ago was (x-5) years Arjun age five years ago was (2x-5)years ATQ, 2X-5=3(X-5) 2X-5=3X-I15 15-5=3x-2x 10=x Shiv present age =x=10years Arjun present age = 2x=2*10=20 years
• 16. Means reducing it to calculate is easily (a)solve: X/2-1/5=x/3+1/4 or x/2-x/3=1/4+1/5 or 3x-2x/6=5+4/20 or x/6=9/20 or x=9*6/20 or x=54/20 or x27/10 = 2.7