For this case we have the following polynomial:
We make the following change of variables:
Rewriting we have
Factoring the second order polynomial we have:
Then, returning the change we have:
Finally, we factor the expression in the parentheses of the second term:
Answer:
the completely factored form is:
The range of a relation is the possible output values, the y-values.
Answer:
(-7, -12)
Step-by-step explanation:
4x-3y=8
5x-2y=-11
Is there any of the like terms can be added and the result will be 0? No, so we have to multiple one OR both of the equations to make that one number do that.
(I will try to remove the y like terms so i will multiple both of them by the opposite so both of the ys will be 6)
2(4x-3y=8)
-3(5x-2y=-11)
8x-6y=16
-15x+6y=33
(now the easy part… cancel the 6s and add the equations)
8x+(-15x)=-7x
16+33=49
-7x=49
(divide 49 by -7)
x=-7
Replace x in any of the equations and you’ll get the y value.
4x-3y=8
4(-7)-3y=8
-28-3y=8
-3y=36
y=12
Threfore, there is one solution which is….. (-7,-12)
Step-by-step explanation:
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