The product of any number and its reciprocal is ' 1 '.
Call the number ' Q '.
Whatever the number is, its reciprocal is 1/Q .
Their product is
( Q ) x ( 1/Q ) = ( Q/Q ) = 1
(x,y)
sub points to see if we get true statemtn
(2,12)
x=2 and y=12
12-5=5(2)
7=10
false
no, (2,12) is not a point
(-2,-5)
x=-2
y=-5
-5-5=5(-2)
-10=-10
true
(-2,-5) is a soltuion
(-2,-5) is the only solution given that works
Answer:
15w - 10x + 30.
Step-by-step explanation:
-5(2x - 3w - 6)
= (-5 * 2x) + (-5 * -3w) + (-5 * -6)
= -10x + 15w + 30
= 15w - 10x + 30.
Hope this helps!
Answer:
Step-by-step explanation:
A system of linear equations is one which may be written in the form
a11x1 + a12x2 + · · · + a1nxn = b1 (1)
a21x1 + a22x2 + · · · + a2nxn = b2 (2)
.
am1x1 + am2x2 + · · · + amnxn = bm (m)
Here, all of the coefficients aij and all of the right hand sides bi are assumed to be known constants. All of the
xi
’s are assumed to be unknowns, that we are to solve for. Note that every left hand side is a sum of terms of
the form constant × x
Solving Linear Systems of Equations
We now introduce, by way of several examples, the systematic procedure for solving systems of linear
equations.
Here is a system of three equations in three unknowns.
x1+ x2 + x3 = 4 (1)
x1+ 2x2 + 3x3 = 9 (2)
2x1+ 3x2 + x3 = 7 (3)
We can reduce the system down to two equations in two unknowns by using the first equation to solve for x1
in terms of x2 and x3
x1 = 4 − x2 − x3 (1’)
1
and substituting this solution into the remaining two equations
(2) (4 − x2 − x3) + 2x2+3x3 = 9 =⇒ x2+2x3 = 5
(3) 2(4 − x2 − x3) + 3x2+ x3 = 7 =⇒ x2− x3 = −1
