The negative solution is x = –1.
Solution:
Let the number be x.
Square of a number = x²
3 times a number = 3x
Difference of square of a number and 4 = 3 times that number
Subtract 3x from both sides of the equation, we get
–3x can be written as x – 4x.
1st bracket have common term x and bracket have –4 common term.
Take common term x + 1 outside.
<em>By zero factor principle, if AB = 0 then A = 0 or B = 0.</em>
x + 1 = 0, x – 4 = 0
x = –1, x = 4
The negative solution is x = –1.
Answer:
The points for the given to linear equations is (5 , - 2) and (5 , - 1)
The points is plotted on the graph shown .
Step-by-step explanation:
Given as :
The two linear equation are
y = x - 1 ...........1
y = x - 6 ...........2
Now, Solving both the linear equations
Put the value of y from eq 2 into eq 1
I.e x - 6 =
x - 1
Or, x +
x = 6 - 1
Or, x = 5
or, x = 5
∴ x = 5
Now, Put the value of x in eq 1
So, y = x - 1
Or, y = × 5 - 1
or, y = - 1
Or, y = - 1 - 1
I.e y = -2
So, For x = 5 , y = - 2
Point is ( ,
) = (5 , - 2)
Again , put the value of x in eq 2
So, y = x - 6
Or, y = × 5 - 6
Or, y = - 6
Or, y = 4 - 6
I.e y = - 2
So, For x = 5 , y = - 2
Point is ( ,
) = (5 , - 2)
Hence, The points for the given to linear equations is (5 , - 2) and (5 , - 2)
The points is plotted on the graph shown . Answer
