Could someone please help me with these systems. I have to use substitution.

1 answer:

8 0

Isolate one term (x or y) in one of the equations

6x+6y=0
6x=-6y
X= -1y

Substitute x=-y into the other equation

-7(-y) + 2y =18

7y + 2y =18

9y=18

Y=2

Substitute y=2 into 6x+6y=0

6x + 6(2) = 0

6x + 12=0

6x= -12

X = -2

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Juliette [100K]

X=-1 y=-1, x=1 y=5, x=3 y=11

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3 years ago

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The temperature fell from 0 Degrees Fahrenheit to 15 and one-half Degrees Fahrenheit below 0 in 5 and three-fourths hours. Wen t

Kryger [21]

Answer:

The correct answer will be:

$-\dfrac{62}{23}$

Step-by-step explanation:

It is given that :

Initial temperature, $T_1 = 0^\circ F$

Final temperature,

$T_2 = -15\dfrac{1}{2}^\circ F\\\Rightarrow T_2 = -\dfrac{15\times 2+1}{2} ^\circ F\\\Rightarrow T_2 = -\dfrac{31}{2} ^\circ F$

Time taken :

$5\dfrac{3}{4}\ hrs = \dfrac{5 \times 4+3}{4}\ hrs = \dfrac{23}{4}\ hrs$

Change in temperature per hour:

$\dfrac{\text{Difference of temperature}}{\text{Total Time Taken}}\\\Rightarrow \dfrac{T_2-T_1}{\text{Total Time Taken}}$

Putting the values of temperatures and time:

$\dfrac{\dfrac{-31}{2}-0}{\dfrac{23}{4}}\\\Rightarrow \dfrac{\dfrac{-31}{2}}{\dfrac{23}{4}}\\\Rightarrow \dfrac{-31 \times 4}{2 \times 23}} \text{---- Error done by Wen at this step}\\\Rightarrow \dfrac{-31 \times 2}{23}}\\\Rightarrow \dfrac{-62}{23}}$