ECUACIONES DE PRIMER GRADO
EJERCICIOS DE ECUACIONES
DE PRIMER GRADO
CON UNA INCÓGNITA
NIVEL 1
Halle el valor de cada incógnita en cada una de las siguientes ecuaciones:
GRUPO A
- 4X=12
- 6X-1=11
- 5X+3=12
- 3X+7=10
- 11X-20=2
- -5X=10+5X
- 2+3X=8
- 10X-12X=12
- 3X=7-37
- 2X+7=X
- 4X-11=2X-3
- X-6=2X+6
- 3X-2X=17-11
- -5X=X+12
- -4X+7=2X+25
- 2X+13=4X+9
- 7X-24=3X-12
- 12-5X=6-3X
- 4y-21-12y=5y+13
- 6a-21+15a= 21a-17
- 34y-54+15y+21y-16=45y+14+23y-18
- α+2α-456-5α-34=21α-345-56α+13α-α
- 0.45t-34+1.45t-2.5=5t-2.9-2.1t+14
- 32.44-2.6t-13.24+t+1.5=4.5t-3.2-t-3.4t+12
- 500-45.23k+12.03k=2.5k+3.5k+4.5k-23
- 99.9-34.6j-100=5.6j+45.1-34j-112
GRUPO C
- 4K-(2K-7)+(5-7K)-(23-11K)=0
- 9R+5(R-2)-7(2R-11)-6(6-4R)=12
- 8(A+15)-3(-2A+1)-7(2A-3)=14(3-3A)
- 4-[-5-(3b-11)]-4[2b-2(5-3b)]=21
- 4β-[-3β-(-2β+15)+3(2-5β)-β]-β=45
- 5L-{-2L+7[-2-5L-(4-7L)+11]-21}=2L
- 2-3λ-2{3-3[λ-2(2λ-11)]-5λ-1}=λ+2(5-λ)-12λ
- 8i-{-[45-31-3(2-3i)-4(5-2i)]-[-2i-4(1-i)]}=0
- 5-{2α-[α-1-(3-2α)-4(3-3α)-α]-3-5α}=2α-8(2-3α)
- 6s-{-4[-2(3-2s)+7]-5[-3(5-2s)]-4(2-3s)}=4
- 2k-{2+[-3-(-2k+2)+7]-7}=-{-[-2-(3-k)]}
- 5-2{-3[1-(2-2x)]-4[5-2(3-3x)]}=x-1
GRUPO D
- (X+5)/2 + (X-2)/3 - (3X-1)/6 = 5/6 RESP: X=-7/2
- (3X-4)/4 - (5X-1)/6 - (3X-2)/24 = -7/12 RESP: X= -4/5
- 3a/4 - (a-3)/7 = a/14 + (2a-1)/28 RESP: a=-1
- (5x-3)/11 = 2x - (3x-2)/22 + (x-1)/2
- 5x/12 - (2-3x)/20 + (5+2x)/24 = 1/60
- (x²-1)/15 + (x-2)/12 - x²/15 = -1/5
- (3x²-2x+7)/6 - (2x²-x-2)/4 - x/8 =17x/24
- (3a-4)/10 + (2a-1)/15 - (4a+1)/12 = 7/30
- X³ - (X-1)³ - 3(X-2)² + 2 = 0
- (3X-2)/7 - 5(x-4)/35 + 1/70 = X/28 RESP: X = -6/5
- (X-1)/4 + (3X-11)/5 - (4X+3)/12 = 7X/60
- (3X+5)/13 + (2X-3)/10 = 5X/26
- (5x+2)² - 5x(5x-4) = 40
- 3x(3x-1)² - (3x-1)³ = 9x² - 5
- (3t-4)/3 + (5t-6)/4 = (2t-9)/12
NIVEL 2
GRUPO A
- 6p-5p(p-2)+p(5p-4)=12
- 45u-2u(5-3u)-4u(2u-11)=-2u(u-7)+6
- -2t(t²+2t-5)-5t(2-4t-3t²)+12=13t(t²+t-2)+3t²-14
- (2x+5)(2x-4)-2(x-5)²-2(x+2)²-25=3x-11
- 3(5-2t)(3-2t)-4[-2t(t-6)-5t²]=10(2t-9)²
- (2t-1)³-(2t+1)³=-6(2t-3)²+4
- (½)(4x²-6x+3)-(¾)(3x²+x-2)+7x(x-7)=2
- (⅕)[4x²-(2x-3)²]+(⅖)[9x²-(2-3x)²]-(⅒)(5x-10)=0
- 6t(2t-7)(2t+7)-24t²(t-5)=30(2t-5)²-40
- (3k+2)(3k-2)(5k+1)(5k-1)-(15k²+1)(15k²-1)+34k²-100=0
- 4x(x+1) - (2x-1)² = 23
- 6t(2t-3)(2t+3)-24t²(t-5)=30(2t-5)²+342
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