The equation y= 2
has one real root and that is x=-1.
What is real roots of the equation?
We are aware that when we resolve a linear or quadratic equation, we always arrive at the value variable of the equation, or, to put it another way, we always locate the equation's solution. This "solution" is what we refer to as the real roots. For instance, when the equation
-7x+12=0 is solved, the actual roots are 3 and 4.
Here given,
=> y = 2
Take y=0 then,
=> 2
=0
=>
=0
=>(x+1)=0
=> x=-1
Hence the given equation has one real root and that is x=-1.
To learn more about real roots refer the below link
brainly.com/question/24147137
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Answer:
true
Step-by-step explanation:
Answer:
<h2>
John has $15 and Alex has $33</h2>
Step-by-step explanation:
a systems of equations can be made from the information on the problem
x+y=48
x=2y+3
since x= 2y+3 substitute 2y+3 in the firat equation to get:
(2y+3)+y=48 -> 3y+3=48 -> 3y=45 -> y=15
plug in 15 for y in the second equation to solve for x
x+15 =48 --> x=33
Answer:
d) 1
Step-by-step explanation:
3a = –2b – 7
We have the point (a, -5) so b = -5
Substituting in
3a = -2(-5) -7
3a = 10-7
3a = 3
Divide each side by 3
3a/3 = 3/3
a =1