बीजगणित का प्रारंभ
Variables: - The quantity represented by letters is called variables. It value is not fixed i.e. x, y, z, a, b, c, l, m, n etc. 2x, 7z, xy, yz, a
अचर:- संख्याओं या अंको से व्यक्त राषि को अचर कहते हैं। अचरों का मान अपरिवर्तनषील होता है। जैसे
Constant: - The quantity represented by numbers is called constant. It value is fixed i.e. 1, 2, 3, …………
पद:- एक चिन्ह वाले मान को पद कहते हैं। जैसे x, y, z, a, b, c, l, m, n, 2x, 7z, xy, yz, abc, x², yz², x²y²z² etc.
Term: - The value of single sign is known as term i.e. x, y, z, a, b, c, l, m, n, 2x, 7z, xy, yz, abc, x², yz², x²y²z² etc.
व्यंजक –> एक या उससे अधिक पदों के संयोग से व्यंजक बनता है। यह कई प्रकार का हो सकता है। जैसे एकपदीय, द्विपदीय , तृपदीय तथा बहुपदीय।
Expression: –> The Expression is formed by adding one or more terms. Thse are of many types like monomials, binomials, trinomials and polynomials.
एकपदीय:- वह व्यंजक जिसमे केवल एक पद होता है। एक पदीय व्यंजक कहलाता है। जैसे x, y, z, xyz, ab²c, 2x, 7z, xy, yz, abc, x², yz², x²y²z² etc.
Monomials: - The Expression that contains only one term is called monomials i.e. x, y, z, xyz, ab²c, 2x, 7z, xy, yz, abc, x², yz², x²y²z² etc.
द्विपदीय:- वह व्यंजक जिसमे केवल दो पद होते हैं। द्विपदीय व्यंजक कहलाता है। जैसे x + y, z – a, 2x + xyz, a + b²c, 2 + x, z – 2, xy + yz, abc + a²bc, x² + yz², x²+y² etc.
Binomials: - The Expression that conatanis only two term is called binomials i.e. x + y, z – a, 2x + xyz, a + b²c, 2 + x, z – 2, xy + yz, abc + a²bc, x² + yz², x²+y² etc.
त्रिपदीय:- वह व्यंजक जिसमे केवल तीन पद होते हैं। त्रिपदीय व्यंजक कहलाता है। जैसे x + y + 2, z – a + b, 2x + xy + z, a + b² + c, a + 2 + x, z – 2 – y, xy + yz – zx, abc + a²bc+ bc, x² + y + z², x²+y² + z² etc.
Trinomials: - The Expression that conatanis only three term is called trinomials i.e. x + y + 2, z – a + b, 2x + xy + z, a + b² + c, a + 2 + x, z – 2 – y, xy + yz – zx, abc + a²bc+ bc, x² + y + z², x²+y² + z² etc.
बहुपद:- वह व्यंजक जिसके गुणज शून्येत्तर (शून्य न हों) तथा घांतें ऋणेत्तर (ऋण न हों) बहुपद कहलाता है। जैसे x + y + 2, z – a + b, 2x + xy + z, a + b² + c, a + 2 + x, z – 2 – y, xy + yz – zx, abc + a²bc+ bc, x² + y + z², x²+y² + z² etc.
Polynomials: - The Expression conataning one or more terms with non zeros coefficen with non negative exponent is known as polynomials i.e. x + y + 2, z – a + b, 2x + xy + z, a + b² + c, a + 2 + x, z – 2 – y, xy + yz – zx, abc + a²bc+ bc, x² + y + z², x²+y² + z² etc.
बहुपद की घात:- किसी बहुपद की उच्चतम धात को बहुपद की घात कहते हैं। 4x + 2 की घात एक तथा
यदि किसी पद में प्रत्येक चार की घात अलग-अलग दी हो तो उस पद या बहुपद की घात उसके पद के चरों की घातों के योग के बराबर होता है।
Power of Polynomials: - Highest power of the polynomials is called the power of the polynomials.
lehdj.k dk ekud :i %& fdlh lehdj.k dks mldh ?kkrksa ds ?kVrs dze esa j[kus ij izkIr :i dks ekud :i dgrs gSaA cgqin x3 – 3x2 – 9x –5 ekud :i esa fy[kk gSA
Standard form of Equation: - Writing of any equation in the order of decreasing powers is known as standard form of equation. The polynomial x3 – 3x2 – 9x –5 is in the standard form.
jsf[kd cgqin %& ?kkr ,d okys cgqin dks jSf[kd cgqin dgrs gSaA tSls
tc blesa ,d gh izdkj ds pj gks rks bls ge ,d pj okyk jSf[kd lehdj.k dgrsa gSA ;fn lehdj.k esa nks pj gks rks mls nks pj okyk cgqin dgrs gSaA
Linear polynomials: - The polynomials of power one is called linear polynomials i.e.
If there are one variable in the equation the it is called Linear equation of one variables if it has two variables then it is called linear equation of two variables.
f}?kkr cgqin%& ?kkr nks okys cgqin dks f}?kkr cgqin dgrs gSaA
Quadratic Polynomials: - The polynomials with power two is known as quadratic polynomials. It can be denoted in the form of
x3 – 2x2 – x + 2, x3 – 3x2 – 9x – 5, x3 + 12x2 + 32x + 20
f=?kkr cgqin %& ?kkr rhu okys cgqin dks f=?kkr cgqin dgrs gSaA bls 
x3 – 2x2 – x + 2, x3 – 3x2 – 9x – 5, x3 + 12x2 + 32x + 20
Cubical Polynomials: - The polynomials with power three is known as cubical polynomials. It can be denoted in the form of
x3 – 2x2 – x + 2, x3 – 3x2 – 9x – 5, x3 + 12x2 + 32x + 20
xq.kkad %& fdlh chth; jkf’k ds lkFk tqMh la[;k dks ml jkf’k dk xq.kkad dgrs gSaA ;g /kukRie rFkk _.kRed gks ldrk gSA /ku dks fcu fpUg ds rFkk _.kRed dks _.k fpUg ds lkFk fy[krs gSaA leh0 
Coefficients: - The numerical factor of the term is called the coefficient. It may be positive or negative. Positive is written without sign and negative written with negative sign. In the equation 
cgqin dk ‘kwU;] ‘kwU;d] xq.ku[kaM ;k gy %& os eku tks lehdj.k esa izfrLFkkfir djus ij mls ‘kwU; esa cny nsrs gSaA cgqin ds ‘kwU;] ‘kwU;d] xq.ku[kaM ;k gy dgykrs gSaA
Zeros, root, and solution of polynomials: - A value of the variable which makes the polynomial zero is called the zero, root or solution of the polynomials.
?;ku jgs fd fdlh cgqin dh ?kkr ds cjkcj mlds gy ;k ‘kwU; gks ldrs gSaA
It is notable that the solutin of the equatin is eqtual to the number of power.
cgqin dk fy[kuk %& dksbZ Hkh f}/kkr cgqin dks 
/;ku jgas fd a,b,c la[;kRed gksrs gSA
Writing of Polynomial: - Any polynomials can be written in the form of 
Note: - a, b and c are written in numerical value.
dHkh dHkh :V ;k oxZewy ;k dsoy pj la[;k,sa Hkh vk tkrh gSa rks ge mUgs bl izdkj le>saA
:V ds fy, dqN fu;e uhps fn;s x;sa tk jgs gSa %&
State which of the following are monomials, binomials and trinomials:
fuEu esa ls ,d inh;] f}inh; rFkk f=inh; O;atdksa dks NkaVks%
4x – 3y
x2
7
4 p2 q – 4 q2 p + r
x + 3,
2y – 5,
3x2,
4xy + 7,
x,
x – 4,
2x + 1,
3x – 2,
17,
– z + 5,
x + y + z,
y + z + 100,
ab – ac,
a + b + c + d,
3xy,
7xyz – 10,
2x + 3y + 7z,
a + b,
4l + 5m,
a + 4,
5 –3xy,
z2 – 4y2,
4x2,
3xy,
–7z,
5xy2,
10y –9,
82a + b + c,
2x + 3y – 5,
x2y – xy2 + y2,
2. Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?
x + y, 1000, x + x2 + x3 + x4, 7 + y + 5x, 2y – 3y2,
2y – 3y2 + 4y3, 4z – 15z2, 5x – 4y + 3xy, pqr, p2q + pq2,
ab + bc + cd + da, 2p + 2q
Classify into monomials, binomials and trinomials.
(i) 4y – 7z (ii) y2 (iii) x + y – xy (iv) 100
(v) ab – a – b (vi) 5 – 3t (vii) 4p2q – 4pq2 (viii) 7mn
(ix) z2 – 3z + 8 (x) a2 + b2 (xi) z2 + z (xii) 1 + x + x2
Look at the following expressions and find like terms:
7x, 14x, –13x, 5x2, 7y, 7xy, –9y2, –9x2, –5yx
Like terms from these are:
(i) 7x, 14x, –13x are like terms.
(ii) 5x2 and –9x2 are like terms.
14. In the following, which pairs contain unlike terms?
3 x , – 7x,
11x, 11y
14xy, –21xy
15ab – 4b
15. Identify like terms in the following:
– xy2, – 7 yx2, – 6 x2 z2, – 18 z2x2, 3x2y, – 5y2x2, 2xy2, 6x2y2,
– 9 x2z2
Identify like terms in the following:
(a) – xy2, – 4yx2, 8x2, 2xy2, 7y,
– 11x2, – 100x, – 11yx, 20x2y,
– 6x2, y, 2xy, 3x
(b) 10pq, 7p, 8q, – p2q2, – 7qp,
– 100q, – 23, 12q2p2, – 5p2, 41, 2405p, 78qp, 13p2q, qp2, 701p2
Find the value of the followings:
fuEu dk eku Kkr djks
x × x
x + x
x – x
x ÷ x
4x × 2x
4x + 2x
4x – 2x
4x ÷ 2x
x × 4
4 × x
x – 4
4 – x
x + 4
4 + x
x ÷ 4
4 ÷ x
x × x × x
x + x + x
y × y × y
y + y + y
Find the value of the followings:
fuEu dk eku Kkr djks
2x × 2x
2x + 2x
2x – 2x
2x ÷ 2x
4x × 2x × x
4x + 2x – 2x
x × 4 x × 5
5 × 4 x × x
4 × x × y
4x × 2x ×5 x
2x + 3x + 2x
2y ×3 y × 2y
3y +2 y + 2y
y × y × y × y
x × x × x × x
chth; O;atdksa ds tksM o ?kVk %&
chth; O;atdksa dks tksMus ds fy, /;ku esa j[ksa fd lcls igys leku inksa dks ;k rks ikl ikl fy[k ysrs gSa ;k fQj Åij uhps dkWye esa fy[k ysrs gSaA
tksMus esa ge fpUgksa dks ugha cnyrs tcfd ?kVkrs oDr nwljs O;atd ds lHkh fpUgksa dks cny nsrs gSaA ;g gesa Å/oZ fof/k esa vklkuh ls le> vk tk;sxk tcfd {ksfrt fof/k esa dks”Bd ds ckgj dk _.k ;k ek;ul fpUg tc iwjs dks”Bd ls xq.kk djrk gS rks og Lo;a gh fpUgksa dks cnyrk pyk tkrk gSA
nks ;k vf/kd leku inksa dk ;ksx ,d leku in gksrk gS] ftldk la[;kRed xq.kkad lHkh leku inksa ds xq.kkadksa ds ;ksx ds cjkcj gksrk gS A
The sum of two or more like terms is a like term with a numerical coefficient equal to the sum of the numerical coefficients of all the like terms.
nks leku inksa dk varj ,d leku in gksrk gS] ftldk la[;kRed xq.kkad nksuksa leku inksa osQ la[;kRed xq.kkadksa osQ varj osQ cjkcj gksrk gSA
The difference between two like terms is a like term with a numerical coefficient equal to the difference between the numerical coefficients of the two like terms.
tksMus o ?kVkusa ds nks fof/k;k¡ gSaA Å/kok/kZj fof/k rFkk {ksfrt fof/k
Å/kok/kZj fof/k (Vertical Method) :- tksM+s o ?kVk;s tkus okys izR;sd O;atd dks ge fofHkUu iafDr;ksa esa fy[krs gSaA ,slk djrs le; ge leku inksa dks ,d nwljs ds Åij&uhps fy[krs gSa vkSj] tSlk uhps n'kkZ;k x;k gSA bl fof/k dks Å/kok/kZj fof/k (Vertical Method) bls dkWye eSFkM+ ds uke ls Hkh tkuk tkrk gSA
Å/kok/kZj fof/k (Vertical Method) bls dkWye eSFkM+ ds uke ls Hkh tkuk tkrk gSA
{ksfrt fof/k (Horizontal Method): - tc O;atdks dks {ksfrt :i esa j[kdj ,d lh/kh js[kk esa tksMk ok ?kVk;k tkrk gS rks bl fof/k dks {ksfrt fof/k dgrs gSaA bls js[kh; eSFkM ds uke ls Hkh tkuk tkrk gSA
Å/kok/kZj fof/k (Vertical Method) bls dksye eSFkM ds uke ls Hkh tkuk tkrk gSA
{ksfrt fof/k (Horizontal Method) bls js[kh; eSFkM ds uke ls Hkh tkuk tkrk gSA
16. Simplify each expression by combining like terms:
12b – 7b – 3b
– x2 + 4x2 – 8x2 + 11x2
2a – (b–a) –b – (a–b)
(x2 + 3x – 2) – (4x – 2x2 – 2)
xy2 – y2 + x2 + xy2 – 4 y2 – x2 – 7
12m2 – 9m + 5m – 4m2 – 7m + 10
21b – 32 + 7b – 20b
– z2 + 13z2 – 5z + 7z3 – 15z
3mn, – 5mn, 8mn, – 4mn
t – 8tz, 3tz – z, z – t
– 7mn + 5, 12mn + 2, 9mn – 8, – 2mn – 3
a + b – 3, b – a + 3, a – b + 3
14x + 10y – 12xy – 13, 18 – 7x – 10y + 8xy, 4xy
5m – 7n, 3n – 4m + 2, 2m – 3mn – 5
4x2y, – 3xy2, –5xy2, 5x2y
17. Add the following: (any three)
a + b – c, b + c – a, c + a – b
3x + 4y – 15z, 6x + 7y, 12y – 7z – 9x
15a + 11b – 13c – 17, 18 – 12c – 7b – 3a
6x – 3y, 3y – 5x + 3z, – x + 2y – 3z
18y2 and 3y2
6ab and –12ab
a + b – c and – a – b – c
3abc–a2–b2 and c2+2a2–b2+abc
x2 - 3xy - 2y2 and 2x2 + 4xy – 5y2
- m2 + 3mn and 3m2 – 3mn + 8
ab – bc, bc – ca, ca – ab
a – b + ab, b – c + bc, c – a + ac
2p2q2 – 3pq + 4, 5 + 7pq – 3p2q2
l2 + m2, m2 + n2, n2 + l2, 2lm + 2mn + 2nl
18. Subtract: (any four)
18y2 from 3y2
6ab from –12ab
a + b – c from – a – b – c
3ab c – a2 – b2 from c2 +2a2– b2 + abc
x2 - 3xy - 2y2 from 2x2 + 4xy – 5y2
- m2 + 3mn from 3m2 – 3mn + 8
b + c – a, c + a – b
3x + 4y – 15z, 12y – 7z – 9x
15a + 11b – 13c – 17, 18 – 12c – 7b – 3a
3y – 5x + 3z, – x + 2y – 3z
4. (a) Subtract 4a – 7ab + 3b + 12 from 12a – 9ab + 5b – 3
(b) Subtract 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + 10xyz
(c) Subtract 4p2q – 3pq + 5pq2 – 8p + 7q – 10 from
18 – 3p – 11q + 5pq – 2pq2 + 5p2q
19. From the sum of 3a – 5b + 3c and 2a + 4b – 5c, subtract 4a – b – c + 3.
20. What should be added to x2 + xy + y2 to obtain 2x2 + 3xy?
21. From the sum of 2x2 + 3xy, – x2 – xy – y2 and xy + 2y2, subtract the sum of 3 x2 – y2 and – x2 + xy + y2.
22. Subtract the sum of 13m – 11n + 9p and – 7p + 3m – 5n from the sum of 6m – 7n – 5p, – 4m + 6p – 9n and 5m – 4n + 3p.
23. If A = 3x2 – 7x + 8, B = x2 + 8x – 3 and C = – 5x2 – 3x + 2, find the value of B – C – A.
24. If a = x – 2, b = y + 2 and c = – x + 2y, show that a + b + c = 3y.
xq.kk djrs oDr fuEu ckrksa dks /;ku esa j[krs gSaA
dks”Bd dk vFkZ xq.kk gksrk gSA
ftl in ij dksbZ fpUg ugha gksrk ml ij tek ;k Iyl dk fu’kku ;k fpUg ekuk tkrk gSA
bl in dh ?kkr] gj rFkk xq.kkad lnSo ,d ekuk fy;k tkrk gSA
lcls igys fpUgksa dh xq.kk djrs gSaA blds fy, fu;e ua0 02 nsa[ksaA
blds mijkar vadksa dh xq.kk djrs gSaA
blds mijkar leku inksa dh xq.kk djrs gSaA leku inksa dh xq.kk djrs oDr leku inksa dh ?kkrkas dks tksM nsrs gSaA
ifjes; ;k fHkUu la[;kvsa ds val dh val esa rFkk gj dh gj ls xq.kk djrs gSaA
vkbZ;s xq.kk djds fuEu lkjf.k;ksa dks iwjk djsaA
1. Find the product of the following pairs of monomials.
fuEu ,d inh; O;atdksa ds xq.kuQy Kkr djksA
4 ×7p
– 4p × 7p
– 4p × 7pq
4p3 × – 3p
4p × 0
(p × q);
(10m × 5n);
(20x2 × 5y2);
(4x × 3x2);
(3mn× 4np)
xy × yz × zx
a× – a2 × a3
2× 4y× 8y2× 16y3
a × 2b × 3c × 6abc
m × – mn × mnp
5a × 3a2 × 7a4
2p × 4q × 8r
xy × 2x2y × 2xy2
xy × 2xy × 2xy
a × 2b × 3c
Find the product
2x (3x + 5xy)
a2 (2ab – 5c)
4p, q + r
ab, a – b
a + b, 7a2b2
a2 – 9, 4a
pq + qr + rp, 0
(a2) × (2a22) × (4a26)
x × x2 × x3 × x4
Multiply the binomials.
(2x + 5) and (4x – 3)
(y – 8) and (3y – 4)
(2.5l – 0.5m) and (2.5l + 0.5m)
(a + 3b) and (x + 5)
(2pq + 3q2) and (3pq – 2q2)
Find the product.
(5 – 2x) (3 + x)
(x + 7y) (7x – y)
(a2 + b) (a + b2)
(p2 – q2) (2p + q)
(x + 3) (x + 3)
(2y + 5) (2y + 5)
(2a – 7) (2a – 7)
(2a – 3b) (2a – 4b)
(2a – 5b) (2a – 7b)
(a2 + b2) (– a2 + b2)
(6x – 7) (6x + 7)
(– a + c) (– a + c)
(x + 3) (x + 7)
(4x + 5) (4x + 1)
(4x – 5) (4x – 1)
(4x + 5) (4x – 1)
(2x + 5y) (2x + 3y)
(2a2 + 9) (2a2 + 5)
(xyz – 4) (xyz – 2)
Find the product.
(5 – 2x) (3 + x + y)
(x + 7y) (7x – y+3)
(a2 + b) (a + b – 5)
(p2 – q2) (2p + q+7)
(x + 3) (x2 + x + 3)
(2y + 5) (2y 2 + y + 5)
(2a – 7) (2a2 + a – 7)
(2a – 3b) (2a – 4b+6)
(2a – 5b) (2a – 7b+1)
(a2 + b2) (5– a2 + b2)
(6x – 7) (6x 2 + x + 7)
(– a + c) (8 – a + c)
(x + 3) (x 2 + x + 7)
(4x + 5) (4x2 +x + 1)
(a + b + c)(a + b – c+ 8)
(a + b + c)(a + b – c)
Divide the followings:
(3x + 5xy) ÷ 2x
(2ab – 5c) ÷ a2
(q + r
ab, a – b
a + b, 7a2b2
a2 – 9, 4a
pq + qr + rp, 0
(a2) × (2a22) × (4a26)
x × x2 × x3 × x4
fuEufyf[kr foHkktu dhft, %
(i) 28x4 ÷ 56x
(ii) –36y3 ÷ 9y2
(iii) 66pq2r3 ÷ 11qr2
(iv) 34x3y3z3 ÷ 51xy2z3
(v) 12a8b8 ÷ (– 6a6b4)
(v) –20x ÷ 10x2
(ii) 7x2y2z2 ÷ 14xyz
2. fn, gq, cgqin dks fn, gq, ,dinh ls Hkkx nhft, %
(5x2 – 6x) ÷ 3x
(3y8 – 4y6 + 5y4) ÷ y4
8(x3y2z2 + x2y3z2 + x2y2z3) ÷ 4x2y2z2
(x3 + 2x2 + 3x) ÷ 2x
(p3q6 – p6q3) ÷ p3q3
Divide (4y3 + 5y2 + 6y) by 2y ls foHkktu
24 (x2yz + xy2z + xyz2) ÷ 8xyz
Divide
from
(7x2 + 14x) ÷ (x + 2)
Divide
, by
.
(y2 + 7y + 10) ÷ (y + 5)
(m2 – 14m – 32) ÷ (m + 2)
(5p2 – 25p + 20) ÷ (p – 1)
Divide
.
Divide
from
.
(07.) gy %&
(08.) gy %& (09.) gy %& |
Divide the following polynomials with x + 1.
fuEu cgqinksa dks x + 1 ls Hkkx djksA
x2 + 3x + 2
x2 – 1
x2 – 2x –3
x3 + x2 + x + 2
x4 + x3 + x2 + x + 1
x4 + 3x3 + 3x2 + x + 1
3. fuEufyf[kr foHkktu dhft, %
(10x – 25) ÷ 5
(10x – 25) ÷ (2x – 5)
10y(6y + 21) ÷ 5(2y + 7)
9x2y2(3z – 24) ÷ 27xy(z – 8)
96abc (3a – 12) (5b – 30) ÷ 144(a – 4) (b – 6)
4. funsZ'kkuqlkj Hkkx nhft, %
5(2x + 1) (3x + 5) ÷ (2x + 1)
26xy(x + 5) (y – 4) ÷ 13x(y – 4)
52pqr (p + q) (q + r) (r + p) ÷ 104pq(q + r) (r + p)
20(y + 4) (y2 + 5y + 3) ÷ 5(y + 4)
x(x + 1) (x + 2) (x + 3) ÷ x(x + 1)
4yz(z2 + 6z – 16) ÷ 2y(z + 8)
5pq(p2 – q2) ÷ 2p(p + q)
12xy(9x2 – 16y2) ÷ 4xy(3x + 4y)
39y3(50y2 – 98) ÷ 26y2(5y + 7)
44(x4 – 5x3 – 24x2) dks 11x (x – 8) ls Hkkx nhft,A
z(5z2 – 80) dks 5z(z + 4) ls Hkkx nhft,A
Equation
In an equation there is always an equality sign. The equality sign shows that the value of the expression to the left of the sign (the left hand side or L.H.S.) is equal to the value of the expression to the right of the sign (the right hand side or R.H.S.).
,d lehdj.k esa] lerk ;k lfedk (equality) ;k cjkcj dk fpUg lnSo gksrk gSA lerk dk fpUg ;g n'kkZrk gS fd bl fpUg ds ckb± vksj ds O;atd ^ck;k¡ i{k* ;k (LHS) dk eku fpUg ds nkb± vksj ds O;atd ^nk;k¡ i{k* ;k (RHS) ds eku ds cjkcj gSA
Equation:- An equation is a condition on a variable such that two expressions in the variable should have equal value.
OR
In short, an equation is a condition on a variable. The condition is that two expressions should have equal value. Note that at least one of the two expressions must contain the variable.
lehdj.k %& ,d lehdj.k] ,d pj ij ,slk izfrca/k gksrk gS ftlesa nksuksa i{kksa esa O;atdksa dk eku cjkcj gksuk pkfg, A
vFkok
laf{kIr :i esa] ,d lehdj.k pj ij ,d izfrca/k gksrk gSA izfrca/k ;g gS fd nksuksa O;atdksa ds eku cjkcj gksus pkfg,A è;ku jgs fd bu nksuksa O;atdksa esa ls de ls de ,d esa pj vo'; gksuk pkfg,A
Roll No. 04 : - An equation remains the same, when the expression on the left and on the right are interchanged. This property is often useful in solving equations.
lehdj.k 04 : - fdlh lehdj.k ds ck,¡ vkSj nk,¡ i{kksa ds O;atdksa dks vkil esa cnyus ij] lehdj.k ogh jgrh gSA ;g xq.k cgq/kk lehdj.kksa dks gy djus esa mi;ksxh jgrk gSA
An equation remains the same if the L.H.S. and the R.H.S. are interchanged.
fdlh lehdj.k ds ck,¡ vkSj nk,¡ i{kksa dks ijLij cnyus ij] lehdj.k ugha cnyrkA
01. Write the following statements in the form of equations:
The sum of three times x and 11 is 32.
If you subtract 5 from 6 times a number, you get 7.
One fourth of m is 3 more than 7.
One third of a number plus 5 is 8.
01. fuEufyf[kr dFkuksa dks lehdj.kksa ds :i esa fyf[k, %
x ds frxqus vkSj 11 dk ;ksx 32 gSA
;fn fdlh la[;k ds 6 xqus esa ls vki 5 ?kVk,¡] rks 7 izkIr gksrk gSA
m dk ,d pkSFkkbZ 7 ls 3 vf/kd gSA
fdlh la[;k ds ,d frgkbZ esa 5 tksM+us ij 8 izkIr gksrk gS A
02. Write equations for the following statements:
The sum of numbers x and 4 is 9.
The difference between y and 2 is 8.
Ten times a is 70
The number b divided by 5 gives 6.
Three fourth of t is 15. (vi) Seven times m plus 7 gets you 77.
One fourth of a number minus 4 gives 4.
If you take away 6 from 6 times y, you get 60.
If you add 3 to one third of z, you get 30.
02. fuEufyf[kr dFkuksa ds fy, lehdj.k nhft, %
la[;kvksa x vkSj 4 dk ;ksx 9 gSA
y esa ls 2 ?kVkus ij 8 izkIr gksrs gSaA
a dk 10 xquk 70 gSA (iv) la[;k b dks 5 ls Hkkx nsus ij 6 izkIr gksrk gSA
t dk rhu&pkSFkkbZ 15 gSA
m dk 7 xquk vkSj 7 dk ;ksxiQy vkidks 77 nsrk gSA
,d la[;k x dh pkSFkkbZ esa ls 4 ?kVkus ij vkidks 4 nsrk gSA
;fn vki y ds 6 xqus esa ls 6 ?kVk,¡] rks vkidks 60 izkIr gksrk gSA
;fn vki z ds ,d&frgkbZ esa 3 tksM+sa] rks vkidks 30 izkIr gksrk gSA
03. Convert the following equations in statement form:
03. fuEufyf[kr lehdj.kksa dks lkekU; dFkuksa ds :i esa cnfy, %
x – 5 = 9
5p = 20
3n + 7 = 1
04. Convert the following equations in statement form:
04. fuEufyf[kr lehdj.kksa dks lkekU; dFkuksa ds :i esa fyf[k, %
p + 4 = 15
m – 7 = 3
2m = 7
3p + 4 = 25
4p – 2 = 18
05. Father’s age is 5 years more than three times Raju’s age. Raju’s father is 44 years old. Set up an equation to find Raju’s age.
05. jktw ds firk dh vk;q jktw dh vk;q ds rhu xqus ls 5 o"kZ vf/kd gSA jktw ds firk dh vk;q 44 o"kZ gSA jktw dh vk;q Kkr djus ds fy,] ,d lehdj.k cukb, A
06. Set up an equation in the following cases:
Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. (Take m to be the number of Parmit’s marbles.)
Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. (Take Laxmi’s age to be y years.)
06. fuEufyf[kr fLFkfr;ksa eas lehdj.k cukb, %
bjQku dgrk gS fd mlds ikl] ijehr ds ikl ftrus daps gSa muds ik¡p xqus ls 7 vf/kd daps gSaA bjQku ds ikl 37 daps gSaA ¼ijehr ds d¡pksa dh la[;k dks m yhft,A½
y{eh ds firk dh vk;q 49 o"kZ gSA mudh vk;q] yM+dh dh vk;q ds rhu xqus ls 4 o"kZ vf/kd gSA ¼y{eh dh vk;q dks y o"kZ yhft,A½
07. Set up an equation in the following cases:
The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87. (Take the lowest score to be l.)
In an isosceles triangle, the vertex angle is twice either base angle. (Let the base angle be b in degrees. Remember that the sum of angles of a triangle is 180 degrees).
07. fuEufyf[kr fLFkfr;ksa eas lehdj.k cukb, %
vè;kfidk crkrh gSa fd mudh d{kk eas ,d fo|kFkhZ }kjk izkIr fd, x, vf/kdre vad] izkIr fd, U;wure vad dk nqxquk /ku 7 gSaA izkIr fd, x, vf/kdre vad 87 gSaA ¼U;wure izkIr fd, x, vadksa dks l yhft,A½
,d lef}ckgq f=Hkqt esa 'kh"kZ dks.k izR;sd vk/kkj dks.k dk nqxquk gSA ¼eku yhft, izR;sd vk/kj dks.k b fMxzh gSA ;kn jf[k, fd f=Hkqt ds rhuksa dks.kksa dk ;ksx 180 fMxzh gksrk gSA½
08. A shopkeeper sells mangoes in two types of boxes, one small and one large. A large box contains as many as 8 small boxes plus 4 loose mangoes. Set up an equation which gives the number of mangoes in each small box. The number of mangoes in a large box is given to be 100.
08. ,d nqdkunkj nks izdkj dh isfV;ksa esa vke csprk gSA ;s isfV;k¡ NksVh vkSj cM+h gSaA ,d cM+h isVh esa 8 NksVh isfV;ksa ds cjkcj vke vkSj 4 [kqys vke vkrs gSaA izR;sd NksVh isVh esa vkeksa dh la[;k crkus okyk ,d lehdj.k cukb,A fn;k gqvk gS fd ,d cM+h isVh esa vkeksa dh la[;k 100 gSA
Rule No. 13: - Thus if we add or subtract the same number to both sides of a balance equation, the balance is undisturbed.
fu;e 13 :-
;fn ge ,d lfedk ds nksuksa i{kksa esa ,d gh la[;k tksM+sa ;k ?kVk,sa rks Hkh og lfedk lR; gksrh gSA
Rule No. 14: - If we multiply or divide both sides of the equation by the same number, the balance is undisturbed.
fu;e 14 :-
;fn ge ,d lfedk ds nksuksa i{kksa dks ,d gh 'kwU;srj (non-zero) la[;k ls xq.kk djsa ;k Hkkx nsa] rks Hkh og lfedk lR; gksrh gSA
Note :- In case of the balanced equation, if we
add the same number to both the sides, or
subtract the same number from both the sides, or
multiply both sides by the same number, or
divide both sides by the same number,
The balance remains undisturbed, i.e., the value of the L.H.S. remains equal to the value of the R.H.S.
uksV%& ,d larqfyr lehdj.k dh fLFkfr esa ;fn ge
nksuksa i{kksa esa ,d gh la[;k tksM+sa ;k
nksuksa i{kksa esa ls ,d gh la[;k ?kVk,¡ ;k
nksuksa i{kksa dks ,d gh la[;k ls xq.kk djsa ;k
nksuksa i{kksa dks ,d gh la[;k ls Hkkx nsa rks larqyu esa dksbZ ifjorZu ugha gksrk gS vFkkZr~ LHS vkSj RHS ds eku leku jgrs gSa A
Transposing : - Changing side is called transposing. While transposing a number, we change its sign. See rule no. 03.
LFkkukiUu %& i{k cnyus dks LFkkukiUu (transposing) djuk dgrs gSaA LFkkukiUu djus esa] la[;k dk fpUg cny tkrk gSA blds fy, fu;e uEcj 03 nsas[ksaA
Note: - Transposing means moving to the other side. Transposition of a number has the same effect as adding same number to (or subtracting the same number from) both sides of the equation. When you transpose a number from one side of the equation to the other side, you change its sign. For example, transposing +3 from the L.H.S. to the R.H.S. in equation x + 3 = 8 gives x = 8 – 3 (= 5). We can carry out the transposition of an expression in the same way as the transposition of a number.
uksV%& LFkkukiUu dk vFkZ gS ,d i{k ls nwljs i{k esa tkukA fdlh la[;k dks LFkkukiUu djuk] la[;k dks nksuksa i{kkas esa tksM+us ;k nksukas i{kksa esa ls ?kVkus ds leku gh gSA tc vki ,d la[;k dks ,d i{k ls nwljs i{k esa LFkkukiUu djrs gSa rks vki mlds fpfUgr dks cny nsrs gSa A mnkgj.kkFkZ] lehdj.k x + 3 = 8 esa + 3 dk LFkkukiUu LHS ls RHS djus ij x = 8 – 3 = 5 izkIr gksrk gS A ge O;atdksa dk Hkh LFkkukiUu mlh fof/k ls djrs gSa tSls ,d la[;k dk LFkkukiUu djrs gSa A
10. Solve the following:
10. fuEu dks ljy djks %
6x = 18
7y = 14
2x = 6
4m = 16
x + 4 = 13
x + 5 = 18
y – 5 = 7
x + 10 = 12
x – 10 = 14
7y – 10 = 0
4y – 16 = 0
5m – 25 = 0
5m + 25 = 0
3x + 9 = 0
2x – 6 = 0
The value of the variable for which the equation is satisfied is called the solution of the equation.
pj dk og eku ftlds fy, lehdj.k larq"V gksrk gS] lehdj.k dk gy dgykrk gSA
11. Solve the following:
11. fuEu dks ljy djks %
2m + 4 = 8
4m – 5 = 11
11 + 2a = 25
12x + 4 = 3x
75 – 3a =35
2x – 3 = 5
7 – 4 y = 3
20x = x + 95
The above property gives a systematic method of solving an equation. We carry out a series of identical mathematical operations on the two sides of the equation in such a way that on one of the sides we get just the variable. The last step is the solution of the equation.
mijksDr xq.kksa }kjk lehdj.k dks pj.k fof/k ls gy fd;k tk ldrk gSA gesa nksuksa i{kksa esa ,d ls vf/d xf.krh; lafØ;k,¡ djuh iM+rh gSa] ftlls fd nksuksa esa ls ,d i{k esa gesa dsoy pj izkIr gksA vafre pj.k lehdj.k dk gy gSA
12. Give first the step you will use to separate the variable and then solve the equation:
12. igys pj dks i`Fkd djus okyk pj.k crkb, vkSj fQj lehdj.k dks gy dhft, %
(a) x – 1 = 0 (b) x + 1 = 0
(c) x – 1 = 5 (d) x + 6 = 2
(e) y – 4 = – 7 (f) y – 4 = 4
(g) y + 4 = 4 (h) y + 4 = – 4
13. Give first the step you will use to separate the variable and then solve the equation:
13. igys pj dks i`Fkd djus ds fy, iz;ksx fd, tkus okys pj.k dks crkb, vkSj fQj
lehdj.k dks gy dhft, %
(a) 3l = 42 (b) 
(c) 
(e) 8y = 36 (f) 
(g) 
14. Give the steps you will use to separate the variable and then solve the equation:
14. pj dks i`Fkd djus ds fy,] tks vki pj.k iz;ksx djsaxs] mls crkb, vkSj fQj lehdj.k dks gy dhft, %
(a) 3n – 2 = 46 (b) 5m + 7 = 17
(c) 
15. Solve the following equations:
15. fuEufyf[kr lehdj.kksa dks gy dhft, %
(a) 10p = 100 (b) 10p + 10 = 100
(c) 
(e) 
(g) 3s + 12 = 0 (h) 3s = 0
(i) 2q = 6 (j) 2q – 6 = 0
(k) 2q + 6 = 0 (l) 2q + 6 = 12
16. Solve the following equations.
16. fuEufyf[kr lehdj.kksa dks gy dhft, %
(a) 
(c) 
(e) 
(g) 
(i) 
17. Solve the following equations.
17. fuEufyf[kr lehdj.kksa dks gy dhft, %
2(x + 4) = 12
3(n – 5) = 21
3(n – 5) = – 21
3 – 2(2 – y ) = 7
– 4(2 – x) = 9
4(2 – x) = 9
4 + 5 (p – 1) = 34
34 – 5(p – 1) = 4
18. Solve the following equations.
18. fuEufyf[kr lehdj.kksa dks gy dhft, %
4 = 5(p – 2)
– 4 = 5(p – 2)
–16 = –5 (2 – p)
10 = 4 + 3(t + 2)
28 = 4 + 3(t + 5)
0 = 16 + 4(m – 6)
5 (3 – x) = 10
3 (2 + x) = 12
3 (x + 2) = 2
3y + 7 = 5 – y
8y – 3 = – 2y – 3
6y – 7 = 2 (y + 3)
10 (x + 2) = 5 (x – 3)
3 (x + 1) = x + 5
19. Solve the following equations.
19. fuEufyf[kr lehdj.kksa dks gy dhft, %
SIMPLE MATH’S
12. The formulae of following things are given then find the other terms of the formulae:
12. fuEu lw=ksa ds mi;ksx djds mlds vU; vO;oksa dks Kkr djksA
Area = Length 🞨 Breadth
A = π r2
V= π r2h
∠A + ∠B + ∠C = 1800
Interest = Amount + Principal
1st No. 🞨 2nd No. = LCM 🞨 HCF
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