<span> 7x+2y=5;13x+14y=-1 </span>Solution :<span><span> {x,y} = {1,-1}</span>
</span>System of Linear Equations entered :<span><span> [1] 7x + 2y = 5
</span><span> [2] 13x + 14y = -1
</span></span>Graphic Representation of the Equations :<span> 2y + 7x = 5 14y + 13x = -1
</span>Solve by Substitution :
// Solve equation [2] for the variable y
<span> [2] 14y = -13x - 1
[2] y = -13x/14 - 1/14</span>
// Plug this in for variable y in equation [1]
<span><span> [1] 7x + 2•(-13x/14-1/14) = 5
</span><span> [1] 36x/7 = 36/7
</span><span> [1] 36x = 36
</span></span>
// Solve equation [1] for the variable x
<span><span> [1] 36x = 36</span>
<span> [1] x = 1</span> </span>
// By now we know this much :
<span><span> x = 1</span>
<span> y = -13x/14-1/14</span></span>
<span>// Use the x value to solve for y
</span>
<span> y = -(13/14)(1)-1/14 = -1 </span>Solution :<span><span> {x,y} = {1,-1}</span>
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Processing ends successfully</span></span>
Answer:
40
Step-by-step explanation:
Let the numbers be 2x and 3x
Then their L.C.M. = 6x
BUT also their LCM=48
6x=48
x=48/6
x=8
So,
First number=2x=2x8=16
Second number=3x=3x8=24
The numbers are 16 and 24.
Hence required sum =(16+24)=40
So none of the options are correct
Step-by-step explanation:
I guess method 1 means to deal with whole factors.
x + 5 = (x - 2)(x + 5)
for (x + 5) <> 0 we can divide both sides by this factor :
1 = x - 2
x = 3
for the second solution we deal with
x + 5 = 0
x = -5
so, for x = -5 and x = 3 both functions deliver the same output, and these are the intersection points.
method 2 : we multiply the expression out and solve it then
x + 5 = (x - 2)(x + 5)
x + 5 = x² + 5x - 2x - 10 = x² + 3x - 10
0 = x² + 2x - 15
the general solution to such a square equation is
x = (-b ± sqrt(b² - 4ac))/(2a)
in our case
a = 1
b = 2
c = -15
x = (-2 ± sqrt(2² - 4×1×-15))/(2×1) =
= (-2 ± sqrt(4 + 60))/2 = (-2 ± sqrt(64))/2 = (-2 ± 8)/2 =
= -1 ± 4
x1 = -1 + 4 = 3
x2 = -1 - 4 = -5
and you get the 2 solutions again. as expected, they are the same as with method 1, of course.
