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Solving Equation with Brackets

Polynomial I - Concepts

Solving Linear Equations without Brackets

Solving Equation with Brackets

Finding Value of an Expression

Polynomial Expression

Degree of a Term

Degree of a Polynomial

Standard Form For a Polynomial

Factor And Remainder Theorem

Relationship of Zeroes With Coefficients of Quadratic Polynomial

Graphical Representation of A Polynomial

Forming a Quadratic Polynomial When Zeroes are Given

Multiplication of Polynomials

Class - 10th IMO Subjects

Concept Explanation

Solving Equation with Brackets

Solving Equation with Brackets:

There are situations in which both the sides of an equation contain both variable (unknown quantity) and constant (numerals). In such cases, we first simplify two sides to their simplest forms and then transpose (shift) terms containing variable on R.H,S to L.H.S and constant terms on L.H.S and R.H.S By transposing a term form one side to other side, we mean changing its sign and carrying it to the other side. In transposition the plus sign of the term changs into minus sign on oher side and vice-versa.

The transposition method involves the following steps:

Step I Obtain the linear euation.

Step II Identify the variable (unknown quantity) and constants (numerals).

Step III Simplify the L.H.S and R.H.S to their simplest forms by removing brackets.

Step IV Transpose all terms containing variable on L.H.S and constant terms on R.H.S Note that the sign of the terms will change in shifting then form L.H.S to R.H.S and vice-versa.

Step V SImplify L.H.S and R.H.S in the simplest form so that each side contains just one term.

Step VI Solve the equation obtaineed in step V by dividing both sides by the coefficient of the variable on L.H.S

Example: Solve: $large frac{x+2}{6}-left [ frac{11-x}{3}-frac{1}{4}right ]=frac{3x-4}{12}$

Solution We have,

$large frac{x+2}{6}-left [ frac{11-x}{3}-frac{1}{4}right ]=frac{3x-4}{12}$

The denominators on two sides of the given equation are 6, 3, 4 and 12.

Their LCM is 12. Multiplying both sides of the given equaton by 12, we get

$large 12left (frac{x+2}{6} right )-12left ( frac{11-x}{3}-frac{1}{4} right )=12left ( frac{3x-4}{12} right )$

$large Rightarrow ;;2(x+2)-4(11-x)+12times frac{1}{4}=(3x-4)$

$large Rightarrow ;;2x+4-44+4x+3=3x-4$

$large Rightarrow ;;6x-37=3x-4$

$large Rightarrow ;;6x-3x=37-4$ [Transposing 3x to LHS and -37 to RHS]

$large Rightarrow ;;3x=33$

$Rightarrow ;;x=frac{33}{3}$ [Dividing both sides by 3]

$Rightarrow ;;x=11$

Check Subtituting x = 11 on both sides of the given equation, we get

$L.H.S.=frac{x+2}{6}-left ( frac{11-x}{3}-frac{1}{4} right )$

$=frac{11+2}{6}-left ( frac{11-11}{3}-frac{1}{4} right )=frac{13}{6}-left ( 0-frac{1}{4} right )=frac{13}{6}+frac{1}{4}=frac{26+3}{12}=frac{29}{12}$

$and ;;R.H.S.=frac{3x-4}{12}=frac{3times 11-4}{12}=frac{33-4}{12}=frac{29}{12}$

Thus, for x = 11, we have L.H.S = R.H.S

Sample Questions

(More Questions for each concept available in Login)

Question : 1

$large dpi{80} 3left ( x+2 right )+2left ( 4x-1 right )=17$
A. $x=frac{13}{19}$	B. $x=frac{22}{11}$	C. $x=frac{13}{11}$	D. $x=frac{16}{4}$
Right Option : C
View Explanation