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alexira [117]
3 years ago
5

Consider the following pair of equations:

Mathematics
1 answer:
Oksana_A [137]3 years ago
5 0
There is a special kind of substitution which some books call it by comparison.

When the two equations are both 
y=some function of x
y=some other function of x, then
we can substitute the second equation into the first giving
some function of x = some other function of x and start solving.

For example,
<span>y = –2x + 8 
y = x – 1
substitute second into first
x-1=-2x+8
isolate x on left,
3x=9
x=3
second step is to substitute x=3 into second equation to get
y=x-1=3-1=2
Therefore the solution is (3,2)</span>
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Help quick please with explain
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Answer:

20= v+9+( -16 )

20= v-( 7 )

27= v

Step-by-step explanation:

Adding a negative is the sane as subtracting. Add like terms. Add 7 to both sides. 27= v

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2 years ago
In the figures, ΔCDA ≈ ΔXYZ
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CD = XY = 26 mm
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Evaluate the expression 2x - 5 when x = -5 <br> C) 5<br> D) 15<br> A)-5<br> B) -15
IRINA_888 [86]

Answer:

B) -15

Step-by-step explanation:

8 0
3 years ago
What is the retail price of a shirt that has a wholesale price of $12 and is marked up by 60 percent?
Soloha48 [4]

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3 years ago
Read 2 more answers
System of Equations <br> Need help<br> Trig
Yanka [14]

Looks like the system is

\begin{cases}-2x+y+6z=1\\3x+2y+5z=16\\7x+3y-4z=11\end{cases}

We can eliminate y by taking

(3x+2y+5z)-2(-2x+y+6z)=16-2(1)

\implies3x+2y+5z+4x-2y-12z=16-2

\implies7x-7z=14

\implies x-z=2

so that z=x-2, and

(7x+3y-4z)-3(-2x+y+6z)=11-3(1)

\implies7x+3y-4z+6x-3y-18z=11-3

\implies13x-22z=8

Substitute z=x-2 into this last equation and solve for x:

13x-22(x-2)=8

\implies13x-22x+44=8

\implies-9x=-36

\implies x=4

Then

z=x-2

\implies z=4-2

\implies z=2

Plug these values into any one of the original equation to solve for y:

-2x+y+6z=1

\implies-2(4)+y+6(2)=1

\implies-8+y+12=1

\implies y=-3

Hence the solution is x = 4, y = -3, and z = 2.

6 0
3 years ago
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