We have a system of equations 2x - 4y = 5 and 6x - 3y = 10.
We want to eliminate one variable when we add up the equations.
If we want to eliminate the x variable, we need to multiply the top, bottom or both of the equations with a number that when the equations are added together will eliminate the x variable.
Multiply the top equation by -3.
-3(2x - 4y) = 5 * -3
-6x + 12y = -15
Now when we add the two equations together, the x's will be eliminated.
-6x + 12y = -15
6x + 6y = 10
18y = -5
The x's were eliminated.
4.75 lbs of water and 0.25 lbs of sugar is needed to make 5% of syrup.
<u>Solution:</u>
Let us assume that x is the required pounds sugar to make 5% syrup
which is 
On solving we get,
![\Rightarrow\frac{x}{5} =0.05 [5\%=\frac{5}{100}]](https://tex.z-dn.net/?f=%5CRightarrow%5Cfrac%7Bx%7D%7B5%7D%20%3D0.05%20%5B5%5C%25%3D%5Cfrac%7B5%7D%7B100%7D%5D)
If the required sugar to make 5% syrup is 0.25 lbs then the remaining quantity in the total sugar syrup would be water. That is,
So, the required amount is 0.25 lbs of sugar and 4.75 lbs of water.
Move all terms not containing
|
5
−
8
x
|
|
5
-
8
x
|
to the right side of the inequality.
Tap for fewer steps...
Add
7
7
to both sides of the inequality.
|
5
−
8
x
|
<
8
+
7
|
5
-
8
x
|
<
8
+
7
Add
8
8
and
7
7
.
|
5
−
8
x
|
<
15
|
5
-
8
x
|
<
15
Remove the absolute value term. This creates a
±
±
on the right side of the inequality because
|
x
|
=
±
x
|
x
|
=
±
x
.
5
−
8
x
<
±
15
5
-
8
x
<
±
15
Set up the positive portion of the
±
±
solution.
5
−
8
x
<
15
5
-
8
x
<
15
Solve the first inequality for
x
x
.
Tap for more steps...
x
>
−
5
4
x
>
-
5
4
Set up the negative portion of the
±
±
solution. When solving the negative portion of an inequality, flip the direction of the inequality sign.
5
−
8
x
>
−
15
5
-
8
x
>
-
15
Solve the second inequality for
x
x
.
Tap for more steps...
x
<
5
2
x
<
5
2
Set up the intersection.
x
>
−
5
4
x
>
-
5
4
and
x
<
5
2
x
<
5
2
Find the intersection between the sets.
−
5
4
<
x
<
5
2
-
5
4
<
x
<
5
2
The result can be shown in multiple forms.
Inequality Form:
−
5
4
<
x
<
5
2
-
5
4
<
x
<
5
2
Interval Notation:
(
−
5
4
,
5
2
)
(
-
5
4
,
5
2
)
The value of c is { c | c ∈ R, c ∉ {-1,0,2,12} }.
c is a number such that it is not in the set {-1,0,2,12} else we end up with a domain value having multiple range values, contradicting with the definition of a function.
