Firstly, let's factorise each equation individually - to do this, find 2 numbers that when summed add to the value of the second term, and when multiplied give the value of the third term.
7 and 12 give us 4 and 3 (4+3=7, 4*3=12) -- 8 and 15 give us 5 and 3 (5+3=8, 5*3=15)
Now we can rewrite these equations as (y+4)(y+3) and (y+5)(y+3) respectively.
Putting this in a fraction: (y+4)(y+3)/(y+5)(y+3) -- We can clearly see that there is a y+3 on both sides of the fraction, and given there are no terms outside of the brackets being multiplied, we can directly cancel.
This gives us our final answer:
(y+4)/(y+5)
Answer:
might be wrong..but i think its all real numbers
Escribir la ecuación estándar de un círculo. Dado un círculo en el plano coordenado, Sal encuentra su ecuación estándar, que es una ecuación de la forma (x-a)²+(y-b)²=r².
Answer:
2
(
1
+
x
) (
1
−
x
) over x
^2
Step-by-step explanation:
if you didn't get that its 2
(
1
+
x
) (
1
−
x
) as the numerator and x
^2 as the denominator
Answer:
x = 136/11
, y = 68/11
Step-by-step explanation:
Solve the following system:
{6 x - y = 68
2 y = x
Hint: | Choose an equation and a variable to solve for.
In the second equation, look to solve for x:
{6 x - y = 68
2 y = x
Hint: | Reverse the equality in 2 y = x in order to isolate x to the left hand side.
2 y = x is equivalent to x = 2 y:
{6 x - y = 68
x = 2 y
Hint: | Perform a substitution.
Substitute x = 2 y into the first equation:
{11 y = 68
x = 2 y
Hint: | Choose an equation and a variable to solve for.
In the first equation, look to solve for y:
{11 y = 68
x = 2 y
Hint: | Solve for y.
Divide both sides by 11:
{y = 68/11
x = 2 y
Hint: | Perform a back substitution.
Substitute y = 68/11 into the second equation:
{y = 68/11
x = 136/11
Hint: | Sort results.
Collect results in alphabetical order:
Answer: {x = 136/11
, y = 68/11
