I need help on these problems. There over solving equations by factoring. Don't get it at all.
1. (Y+5)(Y-7)=0
2. 2n(n-20)=0
3. (2t-3)(3t-2)=0
4. (2u+7)(3u-1)=0
1. (Y+5)(Y-7)=0
if Y = -5
then you have (-5+5) (something) = 0
or 0 * something = 0
so 0 = 0
so Y = -5 IS A SOLUTION
same deal for Y = 7
(something)( 7 - 7 ) = 0
sure enough y = 7 Is a solution
so I claim in general
if (something ) (a x + b) = 0
then
x = -b/a
IS a solution
for example:
2n(n-20) = 0
( 2n + 0) (something ) = 0
so n = -0/2 = 0 IS A SOLUTION
and
something * ( n -20)= 0
if n = -(-20/1) = +20 then (n-20) = 0
then n = + 20 IS A SOLUTION
t = +3/2 and t = +2/3
u = -7/2 and u = +1/3
1.(Y+5)(Y-7)=0
y+5=0 y-7=0
y=0-5 y=0+7
y=-5 y=7
2. 2n(n-20)=0
2n=0 n-20=0
2n/2=0/2 n=0+20
n=0 n=20
3.(2t-3)(3t-2)=0
2t-3=0 3t-2=0
2t=0+3 3t=0+2
2t=3 3t=2
2t/2=3/2 3t/3=2/3
t=3/2 t=2/3
4.(2u+7)(3u-1)=0
Follow above!
Sure! Solving equations by factoring is a method used to find the values of the variable that make the equation true. To solve these problems, we need to factor the given expressions and set each factor equal to zero. Let's go through each problem step by step:
1. (Y+5)(Y-7) = 0
To solve this equation, we set each factor equal to zero:
Y + 5 = 0 or Y - 7 = 0
Now, we solve for Y in each equation:
Y = -5 or Y = 7
So, the solutions to this equation are Y = -5 and Y = 7.
2. 2n(n-20) = 0
Again, set each factor equal to zero:
2n = 0 or n - 20 = 0
Now, solve for n:
n = 0 or n = 20
The solutions to this equation are n = 0 and n = 20.
3. (2t-3)(3t-2) = 0
Set each factor equal to zero:
2t - 3 = 0 or 3t - 2 = 0
Solve for t:
t = 3/2 or t = 2/3
The solutions to this equation are t = 3/2 and t = 2/3.
4. (2u+7)(3u-1) = 0
Set each factor equal to zero:
2u + 7 = 0 or 3u - 1 = 0
Solve for u:
u = -7/2 or u = 1/3
The solutions to this equation are u = -7/2 and u = 1/3.
By factoring the given expressions and setting each factor equal to zero, we can find the solutions to these equations.