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max2010maxim [7]
3 years ago
12

y = 3x + 3 y = x − 1 Explain how you will solve the pair of equations by substitution. Show all the steps and write the solution

in (x, y) form.
Mathematics
1 answer:
Leona [35]3 years ago
5 0
Y = 3x + 3
y = x - 1

As you can see, both equations are set to equal y. This means the right sides of each equation are equal, since y is isolated in both equations. So to solve this particular system of equations for x, set the right sides of both equations equal to each other. After you've done that, you can proceed to solve the equation algebraically for the variable, x.

3x + 3 = x - 1
2x + 3 = -1
2x = -4
x = -2

Negative two is the x-value. To find the y-value, substitute -2 for x into either equation and solve algebraically for y.

y = x - 1
y = -2 - 1
y = -3

The final step is to check all work by plugging both x- and y-values back into both equations.

y = 3x + 3
-3 = 3(-2) + 3
-3 = -6 + 3
-3 = -3  --  This is true

y = x - 1
-3 = -2 - 1
-3 = -3  --  This is true.

Answer:
(-2, -3)
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2-14:

A) 3x+5y=3(3)+5(2)

=9+10

=19

B) x²+y²=3²+2²

=9+4

=13

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=-15+9

=-6

D) xy-1=(2×3)-1

=6-1

=5

E) 2x²-y=2(3)²-2

=18-2

=16

F) 2y-x+5+x²-3y=2(2)-3+5+3²-3(2)

=4-3+5+9-6

=9

2-15: area=y²+xy+xy

=(y²+2xy) un²

perimeter=y+y+y++y+x+y/2+y+x+y/2

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4 0
3 years ago
2. Harry just deposited $1500 into a savings account giving 6% interest compounded quarterly.a) How much will be in the account
Studentka2010 [4]

The formula for compound interest is:

\begin{gathered} A=P(1+\frac{r}{n})^{nt} \\ \text{where,} \\ A=\text{ Final amount} \\ r=\text{ Interest rate} \\ n=\text{ Number of times interest applied per period} \\ t=\text{ Number of time period elapsed} \\ P=\text{ Intial principal balance} \end{gathered}

Given data:

\begin{gathered} P=\text{ \$1500} \\ r=6\text{ \%}=0.06 \\ n=4\text{ times (compounded quarterly)} \end{gathered}

a. After ten years, that is t = 10 years, the amount in the account will be

\begin{gathered} A=1500(1+\frac{0.06}{4})^{4\times10} \\ A=\text{ }1500(1+0.015)^{40} \\ A=\text{ }1500(1.015)^{40} \\ A=\text{ \$2721.03} \end{gathered}

b. After twenty years, that is t = 20 years, the amount in the account will be:

\begin{gathered} A=1500(1+\frac{0.06}{4})^{4\times20} \\ A=1500(1.015)^{4\times20} \\ A=1500(1.015)^{80} \\ A=\text{ \$}4935.99 \end{gathered}

c. The time it takes for Harry's initial account value to double will be:

\begin{gathered} A=2\text{ x initial value = 2 }\times\text{ \$1500 = \$3000} \\ 3000=1500(1.015)^{4t} \\ (1.015)^{4t}=\frac{3000}{1500} \\ (1.015)^{4t}=2 \\ \text{ Find the logarithm of both sides} \\ \ln (1.015)^{4t}=\ln 2 \\ 4t=\frac{\ln 2}{\ln 1.015} \\ 4t=46.56 \\ t=\frac{46.56}{4}=11.64 \end{gathered}

Therefore, the time it takes Harry's initial account to double is approximately 11 years

8 0
1 year ago
Write the following phrase as an algebraic expression:<br><br> 18 less than a number y
zhenek [66]
Y-18
hope this helps!
8 0
3 years ago
Read 2 more answers
Please guys help thank you so much
Svet_ta [14]
First find a common denominator so you can compare them easier. I chose 12. Multiply the numerator and denominator by the number that makes the denominator 12. 2/3 is multiplied by 4 and 2/4 is multiplied by 3. You are left with 8/12 and 6/12. 8/12 is bigger so it looks like
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Hope this helps!

6 0
3 years ago
Which problem can be solved using this equation?
nekit [7.7K]
The 3rd answer, when m is next to a number it means “per” for something like per second or (in this case) per minute, etc...
6 0
3 years ago
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