For
solving system of equations, we can use either substitution where we plug one
equation into the other, or elimination where we combine the equations.
-
Using elimination,
you would to eliminate one variable from both equations, so you automatically would
get one equation with one variable!
- Using
substitution
means you are going to solve one equation for one variable and substitute with
its value in the other equation in order to get also an equation with one
variable.
Let's take an example ...
y+x=2 and y-2x = 1
<span>Using <span>elimination, we need to subtract these two equation; one from the other...
y+x=2
-
y-2x=1
-----------
0+3x=1
then
x=1/3 and then substitute into any equation to get y-value</span></span>
y+x=2
y+1/3 = 2 >>>>>
y=5/3NOW...<span>Using substitution
</span>y+x=2 and y-2x = 1 >>(y=1+2x)
Plug (y=1+2x) into y+x=2 and solve for x
y+x=2
(1+2x) + x =2
1+3x = 2
3x=1
again (and for sure)
x = 1/3plug in x=1/3 into any of the equations above to get y:
y+x=2
y+1/3=2
y=5/3DOne !!!!!!
I hope you got
the idea
If you still need help, just let me know.
4 sq.units
area of a rectangle = length × breadth
length = 4units
breadth = 1units
A = l × b
= 4 × 1
= 4 sq.units
<h3>
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The cross product of the normal vectors of two planes result in a vector parallel to the line of intersection of the two planes.
Corresponding normal vectors of the planes are
<5,-1,-6> and <1,1,1>
We calculate the cross product as a determinant of (i,j,k) and the normal products
i j k
5 -1 -6
1 1 1
=(-1*1-(-6)*1)i -(5*1-(-6)1)j+(5*1-(-1*1))k
=5i-11j+6k
=<5,-11,6>
Check orthogonality with normal vectors using scalar products
(should equal zero if orthogonal)
<5,-11,6>.<5,-1,-6>=25+11-36=0
<5,-11,6>.<1,1,1>=5-11+6=0
Therefore <5,-11,6> is a vector parallel to the line of intersection of the two given planes.
Answer:
the answer is x = 0
Step-by-step explanation:
Cancel equal terms on both sides of the equation:
Divide both side of the equation by 5.
Therefore, the answer is x = 0
