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2 years ago

Solve for x: (3x-2)(x+1)=18x+4

Mathematics

2 answers:

Serga [27]2 years ago

8 0

Answer: x= 7.752
explanation:

DochEvi [55]2 years ago

3 0

Answer:

x = 6 , -1/3

Step-by-step explanation:

Use foil method

(3x -2)(x + 1) = 3x*x + 3x*1 + (-2)*x + (-2)*1

= 3x² + <u>3x - 2x</u> - 2 {Combine like terms}

=3x² + x - 2

(3x -2)(x + 1) = 18x + 4

3x² + x -2 = 18x + 4

Subtract 18x to both sides

3x² + x - 18x - 2 = 4

3x² - 17x - 2 = 4

Subtract 4 from both sides

3x² - 17x -2 -4 = 0

3x² - 17x - 6 = 0

Sum = -17

Product = (-6)*3 = -18

Factors = 1 , (-18) {1*(-18)=-18 ; 1 +(-18) = -17}

3x² - 18x + x - 6 = 0

3x(x - 6) + (x -6) = 0

(x - 6)(3x + 1) = 0

x - 6 = 0 ; 3x + 1 = 0

x = 6 ; 3x = -1

x = -1/3

x = 6 , (-1/3)

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3 years ago

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5 0

3 years ago

Write the standard form of the line that passes through the given points. Include your work in your final answer. Type your answ

Nata [24]

Answer: $5x-3y=44$

Step-by-step explanation:

We know that , the equation of a line that passes through (a,b) and (c,d) is given by :_

$(y-b)=\dfrac{d-b}{c-a}(x-a)$

Standard form of equation of line = $Ax+By=C$

Given points = (7, -3) and (4, -8)

Then, the equation of a line that passes through the points . (7, -3) and (4, -8) is given by -

$(y-(-3))=\dfrac{-8-(-3)}{4-7}(x-7)$

$(y+3)=\dfrac{-8+3}{-3}(x-7)$ [∵ (-)(-)=(+)]

$(y+3)=\dfrac{-5}{-3}(x-7)$

$(y+3)=\dfrac{5}{3}(x-7)$

$3(y+3)=5(x-7)$

$3y+9=5x-35$

$9+35=5x-3y$

$44=5x-3y$

Or $5x-3y=44$

Hence, the required equation :- $5x-3y=44$

4 0

3 years ago

Todea what did one pencil say to another mathematics algebra order of operations answer

Serjik [45]

Part 1

$6- \frac{18-3^2}{4+(2-3)} =6- \frac{18-9}{4+(-1)} \\ \\ =6- \frac{9}{4-1} =6- \frac{9}{3} =6-3 \\ \\ =3$

Part 2

$7-8\div4(3^2-1)=7-8\div4(9-1) \\ \\ =7-8\div4(8)=7-8\div32=7- \frac{1}{2} \\ \\ =6 \frac{1}{2} = \frac{13}{2}$

Part 3

$2-2(2-5)^2+3^2=2-2(-3)^2+9 \\ \\ =2-2(9)+9=2-18+9=-7$

Part 4

$(5^2-9\cdot3)^2-11=(25-27)^2-11 \\ \\ =(-2)^2-11=4-11=-7$

Part 5

$[(2-3)^2-5]\cdot4=[(-1)^2-5]\cdot4 \\ \\ =(1-5)\cdot4=-4\cdot4=-16$

Part 6

$12-11\cdot2+16\div8=12-22+2 \\ \\ =-8$

Part 7

$-5+1\cdot3-(7-2^3)=-5+3-(7-8) \\ \\ =-2-(-1)=-2+1=-1$

Part 8

$8+2\cdot3-14\div7=8+6-2 \\ \\ =12$

Part 9

$(8-2)^2+\frac{1}{4}[4-3(6-10)]=6^2+\frac{1}{4}[4-3(-4)] \\ \\ =36+\frac{1}{4}[4-(-12)]=36+\frac{1}{4}(4+12)=36+\frac{1}{4}(16) \\ \\ =36+4=40$

Part 10

$7+(2-3^2)\cdot5=7+(2-9)\cdot5 \\ \\ =7+(-7)\cdot5=7+(-35)=7-35 \\ \\ =-28$

Part 11

$3-2[2^3+(-1)^3]=3-2[8+(-1)] \\ \\ =3-2(8-1)=3-2(7)=3-14 \\ \\ -11$

Part 12

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8 0

3 years ago