I need help solving these Algebra 1 questions.?

Equation N°1:(y+4)(6y+10)=0

As a multiply is equals to zero one of the parts of the equation must be zero....so lets say y+4=A and 6y+10=B

therefore--> A*B=0, for this to be posible A could be zero so 0*B=0 or B could be zero so A*0=0

so for A=0 we have:

y+4=0 <=> y=-4

and for B=0 we have:

6y+10=0 <=> 6y=-10 (by dividing both parts with 6 we get). y=-10/6

So the solution is: y=-4 or y=-10/6

Equation N°2: y=-11y

This one is easy you can add a (-y) to both parts so we get

y=-11y <=> y - y = -11y - y <=> 0 =-12y (then divide both parts with -12) <=> 0/-12 = (-12/-12)*y <=> y=0

Equation N°3 : 3n +6=11n

in this one first isolate the n's by adding -3n to both parts :

3n + 6 = 11n <=> 3n +6 -3n = 11n -3n <=> 6= 8n

then divide each part with 8:

6=8n <=> 6/8 =8/8*n (simplify 6/8 to 3/4 by multipling and dividing with 2) <=> 3/4 = n => n= 3/4

Equation N°4: (x-3)(2x +1)=22

In this one first multiply (with epimeristiki*) x-3 and 2x+1 by multipling first x and second -3 from x-3 with both 2x and 1 as: x*(2x+1) + (-3)(2x + 1) = [x*2x + x*1]+[-3*2x +(-3)*1] = 2x^2 + x + -6x -3 = 2x^2 -5x -3. so:

(x-3)(2x+1)=22 <=> 2x^2 -5x -3= 22 (add -22 to both parts)

<=> 2x^2 -5x -3 -22 = 22 - 22 <=> 2x^2 -5x -25 = 0

And here we use Diakrinusa

D=b^2 -4*a*c = (-5)^2 -4*2*(-25) = 25 -8*(-25)= 25 + 200 = 225

X1,2=[ -b +- (root)(D)] / 2*a = [-(-5) +- (root)(225)] / 2*2=

=( 5 +- 15) / 4 => x1=(5+15) / 4= 20/4= 5

x2=(5-15)/4=-10/4

In conclusion x1=5 and x2 =-10/4

glad if i helped:)

(y+4)(6y+10)=0 either y+4=0 or 6y+10=0 because the original equation equals zero.

if y+4=0 then y=-4. if 6y+10=0 then y=-5/3.

y=-11y. y=0

3n+6=11n. combine the n's

8n=6 divide both sides by 8

n=6/8=3/4

(x-3)(2x+1)=22 multiply out the brackets

2x^2-5x-25 Use quadratic equation x=(-b+or-sqrt(b^2-4ac))/2a

x=(5+or-sqrt(225))/4 find both solutions for x

x=5 or x=-2

Hope that helped.

(y + 4)(6y + 10) = 0 o

.'. y= -4 or, y= -10/6= -5/3 ['.'ab=0 => a=0 or b=0]

y = -11y

12y=0

y=0

3n + 6 = 11n

8n=6

n= 6/8 =3/4

(x-3)(2x + 1) = 22

2x^2 -5x -3 = 22

2x^2 -5x -25=0

apply sreedhar acharya's formula

x= [5+- √(25+ 200)]/4 = [5+-15]/4

= 5, -10/4= -5/2

1) one of the multiples must = 0, so y must = -4 or -10/6=-5/3

y = - 4 or y = -5/3

2) y must = 0, no other solution

3) (11 - 3)n = 6 = 8n ---> n = 6/8 = 3/4

n = 3/4

4) 2x^2 + x - 6x - 3 - 22 = 0

2x^2 - 5x - 25 = 0

x = ( 5 (+/-) sqrt(225)) / 4

x = (5 (+/-) 15)/4

x = 20/4 = 5 or x = -10/4 = -5/2

x = 5 or x = -5/2

The first one is easy, just make it so either one equals zero. Let's simplify it first, though.

(y + 4) (6y + 10) = 0

(y + 4) (3y + 5) = 0

y = -4, or -5/3

The second one is pretty easy too.

y = -11y

12y = 0

y = 0

The third one's even easier...

3n + 6 = 11n

6 = 8n

n = 3/4

And the last one, just expand everything, collect everything to one side, then solve just like the first question.

(x - 3) (2x + 1) = 22

2x^2 + x - 6x - 3 = 22

2x^2 - 5x - 25 = 0

(2x + 5) (x - 5) = 0

x = -5/2 or 5

1st one. multiply (y+4) and (6y+10) you'll get an equation in quadratic..solve and you'll get 2 answers.

2nd. bring RHS to LHS= y+11y=0...=12y=0 or just y=0.

3rd. bring 3n to RHS(right hand side).=.. 6=11n-3n..=> 6=8n..=> n=6/8 or n= 3/4.

4th. multiply (x-3) and (2x+1)..you'll get quadratic and solve..again you'll get 2 answers

Step a million: Your equation will ultimately take the variety of Ax +through = C starting up with y - y1 = m(x - x1) the position x = 5, y = -2 and m = -3 yields: y - (-2) = -3(x-5) y + 2 = -3x + 15 which simplifies to y = -3x + 13. Rewriting through including -3x to each and every aspect of equation yields: 3x + y = 13 that's now in usual variety desire this helps

1.y=-4,y=-5/3

2.??

3.n=3/4

4.??