The answer is 1
For the line y = 1 - 3x, create a table to show the values of x and y where x is from -2 to 2.
For the line y = 1 - 3x the values of x and y are given below:
Given,
y = 1 - 3x
putting x = -2
y = 1 - 3x
y = 1 - 3(-2)
y = 7
Putting x = -1
y = 1 - 3(-1)
y = 4
putting x = 0
y = 1 - 3(0)
y = 1
Now putting x = 1
y = 1 - 3(1)
y = -2
putting x = 2
y = 1 - 3(2)
y = -5
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The question is in the ss
An algebraic expression that is equivalent to the given expression
(2c³)⁵(4c⁴) / (8c⁶)² is equal to 2c⁷.
As given in the question,
Given algebraic expression is equal to :
(2c³)⁵(4c⁴) / (8c⁶)²
Law of indices:
mᵃ × mᵇ = mᵃ⁺ᵇ
mᵃ ÷ mᵇ = mᵃ⁻ᵇ
To get equivalent expression simplify the given expression
(2c³)⁵(4c⁴) / (8c⁶)² using law of indices:
(2c³)⁵(4c⁴) / (8c⁶)²
= ( 2⁵c¹⁵) ( 2²c⁴) / ( 2³c⁶)²
=( 2⁵c¹⁵) ( 2²c⁴) / ( 2⁶c¹²)
= 2⁵⁺²⁻⁶ c¹⁵⁺⁴⁻¹²
= 2c⁷
Therefore, an algebraic expression that is equivalent to the given expression (2c³)⁵(4c⁴) / (8c⁶)² is equal to 2c⁷.
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What is the ones digit in the number 2 to the 2056 Power?Hint: start with smaller exponents to find the pattern
Follow the steps below to find the ones digit in the number:
[tex]\begin{gathered} 2^{2056} \\ \text{The base = 2} \\ \text{The power = 2056} \end{gathered}[/tex]step 1: Identify the unit digit in the base - the unit digit is 2.
step 2:
If the power or exponent is exactly divisible by 4
case 1: The unit digit of the number is 6 if the unit digit in the base is any of 2, 4, 6, 8.
case 2: The unit digit of the number is 1 if the unit digit in the base is 3, 7, 9.
In this case, the power 2056 is exactly divisible by 4 and the unit digit in the base is 2.
Therefore, the ones digit is 6
What’s the correct answer answer asap i need help can somebody answer this question?
il give you brainlist
Answer:
Virtuosity on the piano
Answer: a
Step-by-step explanation:
Maya is playing golf. On her first two holes she scored one under par then six over par. Find her score after the first two holes.
If Maya is playing golf in which on her first two holes she has one under par then six over par. Her score after the first is 5 over per.
Score after the first two holesFirst is to analyze the given the information given
What she got on the first hole = one under par = -1
What she got on her second hole = six over par = 6
Now let find or determine her score
Score = -1 + 6
Score = 5 over par
Therefore we can conclude that her score is 5 over per.
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Please see attachment below Is the following an example of theoretical probability or empirical probability? A fisherman notes that eight out of ten times that he uses a certain lure to catch a fish within an hour. He concludes that the probability that the lure wil catch a fish on his next fishing trip is about 80%. Is it empirical or theoretical?
From the information provided, we know that the fisherman has already conducted some experiments and from these events he was able to conclude the probability of catching a fish on his next fishing trip. This is an example of EMPIRICAL PROBABILITY.
This is probability calculated after the event has occured.
ANSWER:
Empirical
how many many eighths are in 8/9
There are 1/9 eighths in 8/9
whats the average for 62,81,72,60,100,79
Answer:
75.66
Step-by-step explanation:
62, 81, 72, 60, 100, 79
To find the average, add all the numbers together then divide by how many numbers there are.
62 + 81 + 72 + 60 + 100 + 79 = 454
454 ÷ 6 = 75.66 or rounded 76
I hope this helps!
A girl is pedaling her bicycle at a speed of 0.10 km/min. How far will she travel in three hours?
Answer:
18 km
Step-by-step explanation:
.1 km/min * 60 min/hr * 3 hr = 18 km
Answer: 18 km
Step-by-step explanation:
First, you know that 60 minutes equal one hour so to get a km/hr ratio you need to multiply the 0.10km by 60 and the minute (1) to get 6(km)/1(hr), then you multiply the 6km by 3 as well as the 1hr by 3 since we need to get to three hours. With this you get 18(km)/3(hr).
On a flight from Chicago, Illinois, to Denver, Colorado,
the typical cruising altitude of a passenger jet is approximately
38,000 feet. As the flight departs from O'Hare International Airport,
the jet begins to ascend. At 4.3 miles away from the airport, the jet
is at an altitude of 4200 feet. The jet reaches an altitude of 36,700
feet at 167.5 miles away from the airport.
Determine the slope of ascent of the flight given that 5280 feet = 1 mile.
Round your answer to the nearest thousandth.
The slope of ascent of the flight is 0.04
How to calculate the slope of ascent?From the question, we have the following parameters that can be used in our computation:
Typical cruising altitude = 38,000 feetInitial point of ascend = 4.3 miles at 4,200 feetAnother point of ascend = 167.5 miles at 36,700 feetThe above parameters can be represented as
(x, y) = (4.3, 4200) and (167.5, 36,700)
The slope is then calculated as
Slope = (y₂ - y₁)/(x₂ - x₁)
Where x and y are defined above
So, we have
Slope = (36700 - 4200)/(167.5 - 4.3)
Evaluate
Slope = 199.14 feet per mile
Recall that 5280 feet = 1 mile.
So, we have
Slope = 199.14 feet/5280 feet
Evaluate
Slope = 0.04
Hence, the required slope is 0.04
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1. Given d = 25,01 = 15°, and O2 = 40°. Find x and h. Show work. Round to 2 decimal places.0,
Let's check first this right triangle from the figure:
We have, using tangent on the angle:
[tex]\begin{gathered} \tan (15^o)=\frac{h}{25+x} \\ \Rightarrow h=0.268(25+x) \end{gathered}[/tex]And then, looking at the other right triangle:
Doing the same as before, we have that:
[tex]\begin{gathered} \tan (40)\text{ = }\frac{h}{x} \\ \Rightarrow h=0.839x \end{gathered}[/tex]Now, we can use both equations of h to solve for x:
[tex]\begin{gathered} 0.268(25+x)=0.839x \\ \Rightarrow6.7+0.268x=0.839x \\ \Rightarrow6.7=0.571x\Rightarrow x=\frac{6.7}{0.571}=11.73 \end{gathered}[/tex]Which we can finally use to find h in any equation:
[tex]\begin{gathered} h=0.268(25+11.73)=9.843 \\ or \\ h=0.839\cdot11.73=9.841 \end{gathered}[/tex]What is 75.6 divided by 4
Answer:
18.9
Step-by-step explanation:
Calculators, haha
Colin just travelled across Ontario on a road trip. He bought some skis in Blue Mountain for
$879.95 plus tax, a boom box in Muskoka for $145.58 including taxes, a souvenir in Niagara Falls
for $99.97 plus tax, and some maple syrup in Toronto for $45.14 including tax. Overall, how
much HST did Colin pay on his trip? Answer should be rounded off to whole number.
The Harmonized Sales Tax paid by Colin is $152.
What is Harmonized Sales Tax?Almost all buyers of taxable supplies of goods and services must pay the GST/HST (other than zero-rated supplies). Indians, Indian bands, and entities with band authority, however, are occasionally exempt from having to pay the GST/HST on taxable supplies. In Ontario, the HST (Harmonized Sales Tax) is 13%. Ontario offers a point-of-sale rebate for the 8% provincial HST exemption on a selection of goods.So, the amount of HST Colin pays:
The total amount of items brought:
Skis: $879.95Boom box: $145.58Souvenir: $99.97Maple syrup: $45.14Total amount: $1,170.64
We know that HST is 13% of the amount, then:
$1,170.64/100 × 13 = 152.1832Rounding off: $152Therefore, the Harmonized Sales Tax paid by Colin is $152.
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xx yyy
121212 363636
181818 545454
252525 757575
Find the constant of proportionality (r)(r)left parenthesis, r, right parenthesis in the equation y=rxy=rxy, equals, r, x.
The constant of proportionality for the given set is 3.
Given,
The set;
x y
12 36
18 54
25 75
We have to find the constant of proportionality;
Here,
y = x × r
Where r is the constant of proportionality.
Lets see;
Case 1; 12 and 36
36 = 12 × r
r = 36/12
r = 3
Case 2; 18 and 54
54 = 18 × r
r = 54/18
r = 3
Case 3; 25 and 75
75 = 25 × r
r = 75/25
r = 3
Therefore,
The constant of proportionality for the given set is 3.
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What is the slope of a line perpendicular to the line whose equation is x+3y=-15. Fully simplify your answer.
ANSWER
The slope of the line perpendicular line of the equation is 3
STEP-BY-STEP EXPLANATION:
What to find? The slope of a line perpendicular to the line whose equation is x + 3y = -15
Given the equation
x + 3y = -15
The slope-intercept form of an equation is given as
[tex]y\text{ = mx + b}[/tex]Where m = slope of the line
y = the intercept of the y-axis
The next step is to re-arrange the above equation in the slope-intercept format
[tex]\begin{gathered} \text{Given the equation of a straight line as} \\ x\text{ + 3y = -15} \\ \text{Isolate 3y by substracting x from both sides} \\ x\text{ - x + 3y = -15 - x} \\ 3y\text{ = -x - 15} \\ \text{Divide through by 3} \\ \frac{3y}{3}\text{ = }\frac{-1}{3}x\text{ -}\frac{15}{3} \\ y\text{ = }\frac{-1}{3}x\text{ - 5} \\ \text{Hence, the slope}-\text{intercept form of the above equation is given as} \\ y\text{ = }\frac{-1}{3}x\text{ - 5} \end{gathered}[/tex]NB: That the two lines are perpendicular to each other
From y = mx + b
m = -1/3
The slope of the equation
[tex]\begin{gathered} \text{ For two perpendicular lines, we can calculate the slope as follows} \\ m_1\cdot m_2\text{ =- 1} \\ \text{where m}_1\text{ = }\frac{-1}{3} \\ \frac{-1}{3}\cdot m_2\text{ = -1} \\ \frac{-1\cdot m_2}{3}=\text{ -1} \\ \text{Cross multiply} \\ -m_2\text{ = -1 }\cdot\text{ 3} \\ -m_2\text{ = -3} \\ \text{Divide through by -1} \\ \frac{-m_2}{-1}\text{ = }\frac{-3}{-1} \\ m_2\text{ = }3 \\ \text{Hence, the slope of the perpendicular line to the equation is 3} \end{gathered}[/tex]Margaret drove to a business appointment at 70 mph. Her average speed on the return trip was 60 mph. The return took ⅓ hr longer because of heavy traffic. How far did she travel to the appointment ?
Let d and t be the distance and the time it takes for Margaret to get the place of the appointment.
d is the product of the speed when she was going to the appointment times the time (t):
[tex]d=70t[/tex]d is also the product of the speed when she was going back home times t+1/3:
[tex]d=60(t+\frac{1}{3})[/tex]Make both d equal and find the value of t:
[tex]\begin{gathered} 70t=60(t+\frac{1}{3}) \\ 70t=60t+20 \\ 10t=20 \\ t=\frac{20}{10} \\ t=2 \end{gathered}[/tex]Use this value of t to find d:
[tex]\begin{gathered} d=70(2) \\ d=140 \end{gathered}[/tex]According to this, she traveled 140 miles to the appointment.
To pass a course with a B grade a student must have an average of 80 or greater. The students grades on three test are 80, 81 and 70. Find what score the student must get on the next test to get a B average or better.
Given that the student's grades on three tests are 80, 81 and 70
Let the student's grade on the fourth test be X
In order to pass the course, the student must have an average of 80 or greater.
Hence,
[tex]\begin{gathered} \frac{80+81+70+X}{4}\ge80 \\ \frac{231+X}{4}\ge80 \\ 231+X\ge320 \\ X\ge320-231 \\ X\ge89 \end{gathered}[/tex]The score the student must get on the next test to get a B average or better is greater than or equal to 89
A map is drawn using a scale of 1 inch : 4 meters. If the length of a paved path on a map is 16 inches, then find the actual length of the paved path.
Nana, this is the solution to the exercise:
Scale 1 inch : 4 meters
For solving this exercise, we will use the Direct Rule of Three, as follows:
Length on the map (inches) - Actual length (meters)
______________________________________________
1 4
16 x
_______________________________________________
1 * x = 16 * 4
x = 64
The actual length of the paved path is 64 meters
A local newspaper charges $13 for each of the first four lines of a classified ad, and $7.50 foreach additional line. Express the cost of a -line ad, c(x), as a piecewise function
The cost is the basic cost increased by the product of the number of extra lines and the price per additional line, then:
Let x be the number of additional lines
c(x)=13+7.50x
If 4–(x – 4) = 1 over 16 what is the value of x? (2 points)
–2
–6
6
2
Answer: these are the answers that i got -2 for x =10 -6= 14 6=2 and 2= 6
Step-by-step explanation:
Find the percent change and tell whether it is a percent decrease orincrease.Original amount: 30End amount: 45
Step 1
Since the end amount is greater than the original amount, it is, therefore, percentage increase
Step 2
Calculate the percentage increase
The formula is given by
[tex]\begin{gathered} \text{Percentage increase =}\frac{Increase}{\text{Original}}\times100 \\ \text{Increase}=\text{ }45\text{ - 30 = 15} \\ \end{gathered}[/tex][tex]\text{Percentage increase =}\frac{15}{30}\times100=\text{ 50 percent}[/tex]The answer is therefore 50%
Determine the values of x in the equation x^2 = 81. Please help
x = −9
x = ±9
x = ±40.5
x = 40.5
The value of x in the quadratic equation is ±9.
As equation is [tex]X^{2}[/tex] = 81
Taking square root on both sides
⇒ [tex]\sqrt{x^{2} } = \sqrt{81}[/tex]
⇒ X = ± 9
∴ The value of X is ± 9
The roots of the given quadratic equation are +9 and -9.
How to solve a quadratic equation?An equation of degree 2 containing a single variable is known as the quadratic equation. . Its general form is [tex]Ax^{2}+ Bx^{} + C[/tex], where variables are x and constants are a, b, and c.
Examples are [tex]X^{2}[/tex] = ± 3, [tex]X^{2}[/tex] = ±9, [tex]X^{2}[/tex]=4 or any real number.
Some other examples are [tex]x^{2} + 2X+2[/tex], etc.
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The price-demand and cost functions for the production of microwaves are given asP= 180 - q/50and C(q) = 72000 + 110g,where q is the number of microwaves that can be sold at a price of p dollars per unit and C(q) is the total cost (in dollars) of producing q units.(C) Find the marginal revenue function in terms of q.R'(q) =(D) Evaluate the marginal revenue function at q=1100.R'(1100) =(E) Find the profit function in terms of q.P(q)(F) Evaluate the marginal proft function at q = 1100.P'(1100)
Answer:
A)
[tex]\begin{equation*} C^{\prime}(q)=110 \end{equation*}[/tex]B)
[tex]\begin{equation*} R(q)=180q-\frac{q^2}{50} \end{equation*}[/tex]C)
[tex]\begin{equation*} R^{\prime}(q)=180-\frac{q}{25} \end{equation*}[/tex]D)
[tex]\begin{equation*} R^{\prime}(1100)=136 \end{equation*}[/tex]E)
[tex]\begin{equation*} P(q)=-\frac{q^2}{50}+70q-72000 \end{equation*}[/tex]F)
[tex]\begin{equation*} P^{\prime}(1100)=26 \end{equation*}[/tex]Explanation:
Given:
[tex]\begin{gathered} p=180-\frac{q}{50} \\ C(q)=72000+110q \end{gathered}[/tex]where q = the number of microwaves that can be sold at a price of p dollars per unit
C(q) = the total cost (in dollars) of producing q units.
A) To find the marginal cost, C'(q), we'll go ahead and take the derivative of the total cost as seen below;
[tex]\begin{gathered} C(q)=72000+110q \\ C^{\prime}(q)=0+110 \\ \therefore C^{\prime}(q)=110 \end{gathered}[/tex]So the marginal cost, C'(q) = 110
B) We'll go ahead and determine the revenue function, R(q), by multiplying the price, p, by the quantity, q, as seen below;
[tex]\begin{gathered} R(q)=p*q=(180-\frac{q}{50})q=180q-\frac{q^2}{50} \\ \therefore R(q)=180q-\frac{q^2}{50} \end{gathered}[/tex]C) We'll go ahead and determine the marginal revenue function, R'(q), by taking the derivative of the revenue function, R(q);
[tex]\begin{gathered} \begin{equation*} R(q)=180q-\frac{q^2}{50} \end{equation*} \\ R^{\prime}(q)=180-\frac{2q^{2-1}}{50}=180-\frac{q}{25} \\ \therefore R^{\prime}(q)=180-\frac{q}{25} \end{gathered}[/tex]D) To evaluate the marginal revenue function at q = 1100, all we need to do is substitute the q with 1100 in R'(q) and simplify;
[tex]\begin{gathered} \begin{equation*} R^{\prime}(q)=180-\frac{q}{25} \end{equation*} \\ R^{\prime}(1100)=180-\frac{1100}{25}=180-44=136 \\ \therefore R^{\prime}(1100)=136 \end{gathered}[/tex]Therefore, R'(1100) is 136
E) To find the profit function, P(q), we have to subtract the total cost, C(q), from the revenue cost, R(q);
[tex]\begin{gathered} P(q)=R(q)-C(q) \\ =(180q-\frac{q^2}{50})-(72,000+110q) \\ =180q-\frac{q^2}{50}-72000-110q \\ =-\frac{q^2}{50}+180q-110q-72000 \\ =-\frac{q^2}{50}+70q-72000 \\ \therefore P(q)=-\frac{q^2}{50}+70q-72000 \end{gathered}[/tex]F) To Evaluate the marginal profit function at q = 1100, we have to first determine the marginal profit, P'(q), by taking the derivative of the profit function, P(x);
[tex]\begin{gathered} \begin{equation*} P(q)=-\frac{q^2}{50}+70q-72000 \end{equation*} \\ P^{\prime}(q)=-\frac{2q^{2-1}}{50}+70(1*q^{1-1})-0=-2q^{50}+70q^0=-\frac{q}{25}+70 \\ \therefore P^{\prime}(q)=-\frac{q}{25}+70 \end{gathered}[/tex]We can now go ahead and find P'(1100) as seen below;
[tex]\begin{gathered} P^{\prime}(1100)=-\frac{1100}{25}+70=-44+70=26 \\ \therefore P^{\prime}(1100)=26 \end{gathered}[/tex]So P'(1100) = 26
Which combination could the coach buy?Click on the correct answer.Remember, the coach1) needs to buy at least 4 bats and atleast 8 balls2) cannot spend more than $1206 bats, 10 balls12-3 bats, 10 balls84 bats, 9 balls-4-110T14812x24y2815x + 5y = 120
It is important to know that the yellow area represents all the possible solutions because there all the areas intercept. So, among the options, the correct one is 4 bats, 9 balls because it represents (3,10) which is inside the yellow region.
Hence, the answer is option 3.Hello, just want to check my answers at Part B. Thanks!
SOLUTION
The given functions are:
[tex]f(x)=(x-2)(x-1)(x-1),g(x)=\sqrt[3]{x}-2[/tex]Notice that when x=0
[tex]\begin{gathered} f(0)=(0-2)(0-1)(0-1) \\ f(0)=-2 \end{gathered}[/tex]Also
[tex]\begin{gathered} g(0)=\sqrt[3]{0}-2 \\ g(0)=-2 \end{gathered}[/tex]This shows that f(0)=g(0)
Hence there are no breakes in the domain of h(x)
Joe drinks 0.55 L of milk every day. How much milk does he drink in 5 days? Write your answer in milliliters.
Given that
Joe drinks 0.55 l of milk every day and we have to find the milk he can drink in 5 days.
Explanation -
Let the given condition be represented as
1 day = 0.55 L milk
So multiplying by 5 on both sides we have
1 x 5 days = 0.55 x 5 L milk
5 days = 2.75 L milk
And conversion is
1L = 1000 ml
So multiplying by 2.75 in the above equation we have
2.75L = 2.75 x 1000 ml = 2750 ml
So Joe will drink 2750 ml of milk in 5 days.
Hence the final answer is 2750 ml.a man paddles his canoe at a rate of 12 kilometers per hour. write this rate as meters per minute
Answer
12 kilometers per hour = 200 meters per minute
Explanation
In order to convert 12 kilometers per hour into meters per minute, we need to note that
1 kilometer = 1000 meters
1 hour = 60 minutes
[tex]\begin{gathered} 12\frac{kilometers}{hour} \\ We\text{ ne}ed\text{ to note that } \\ \frac{1000\text{ meters}}{1\text{ kilometers}}=1 \\ \frac{1\text{ hour}}{60\text{ minutes}}=1 \\ 12\frac{\text{kilometers}}{\text{hour}}\times1\times1 \\ =12\frac{\text{kilometers}}{\text{hour}}\times\frac{1000\text{ meters}}{1\text{ kilometer}}\times\frac{1\text{ hour}}{60\text{ minutes}} \\ =\frac{12\times1000\times1}{1\times60}\frac{meters}{\min ute} \\ =200\frac{\text{meters}}{\text{minute}} \end{gathered}[/tex]Hope this Helps!!!
When the engine falls out of Rhonda's old car, it's time to shop for something newer. She is hoping to keep her monthly payment at $140, and a loan will be 5% simple interest for 48 months with a $1000 down payment. Under these conditions, the most expensive car that Rhonda can afford is $6600.00. A salesman tries to convince Rhonda that she would be able to get a much better car if she raises the payment by just $30 per month.
The maximum value increase along with the down payment is $1200.
What is Down Payment?A down payment, sometimes known as a deposit in British English, is an upfront partial payment made when purchasing expensive goods or services like a home or car. At the moment the transaction is completed, it is often paid in cash or an equivalent. The remainder of the payment must then be financed using some kind of loan.
We need to calculate the value by which the price increases if she does raise the payment by $30
So, the total payment made by her = $140 +$30 = $170
This payment is made for 48 months at 5% simple interest
Total payments made= $170 × 48 = $8160
Let P be the total car value
P + SI = 8160
P + PTR = 8160
P(1+TR) = 8160
P = 8160/ (1+0.05 × 48 /12)
P= $6800
The maximum price that can be done is $6800 + down payment
hence the maximum price is $7800
Hence, the maximum value increase =$7800 - $6600= $1200
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Question: When the engine falls out of Rhonda's old car, it's time to shop for something newer. She is hoping to keep her monthly payment at $140, and a loan will be 5% simple interest for 48 months with a $1000 down payment. Under these conditions, the most expensive car that Rhonda can afford is $6600.00. A salesman tries to convince Rhonda that she would be able to get a much better car if she raises the payment by just $30 per month. What is the maximum value increase along with the down payment?
Someone please help I don’t understand this at all would really appreciate some help !!!!
The results of the operations of functions evaluated at are listed below:
Addition - (f + g) (- 2) = 9Subtraction - (f - g) (- 2) = 3Multiplication - (f · g) (- 2) = 18Division - (f / g) (- 2) = 2 / 3How to evaluate operations of functions
In this problem we four cases of operations between two functions, a quadratic equation f(x) and a linear equation g(x). There are four operations:
Addition - f (x) + g (x) = (f + g) (x)
Subtraction - f (x) - g(x) = (f - g) (x)
Multiplication - f (x) · g (x) = f · g (x)
Division - f (x) / g (x) = (f / g) (x)
If we know that f(x) = x² - x and g(x) = 3 · x + 9, then the compositions of functions evaluated at x = - 2 are, respectively:
(f + g) (- 2) = (- 2)² - (- 2) + 3 · (- 2) + 9
(f + g) (- 2) = 4 + 2 - 6 + 9
(f + g) (- 2) = 9
(f - g) (- 2) = (- 2)² - (- 2) - 3 · (- 2) - 9
(f - g) (- 2) = 4 + 2 + 6 - 9
(f - g) (- 2) = 3
(f · g) (- 2) = [(- 2)² - (- 2)] · [3 · (- 2) + 9]
(f · g) (- 2) = 6 · 3
(f · g) (- 2) = 18
(f / g) (- 2) = [(- 2)² - 2] / [3 · (- 2) + 9]
(f / g) (- 2) = 2 / 3
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I need help on question 10.The solution of the system below is:3x + 2y = 143x – 2y = 10
Given the system of equations:
[tex]\begin{cases}3x+2y=14 \\ 3x-2y=10\end{cases}[/tex]We'll multiply the second equation by -1 and then add both equations up:
[tex]\begin{gathered} \begin{cases}3x+2y=14 \\ 3x-2y=10\end{cases}\rightarrow\begin{cases}3x+2y=14 \\ -3x+2y=-10\end{cases} \\ \\ \rightarrow4y=4 \end{gathered}[/tex]Solving the resulting equation for y ,
[tex]\begin{gathered} 4y=4\rightarrow y=\frac{4}{4} \\ \\ \Rightarrow y=1 \end{gathered}[/tex]We'll plug in this y-value in the first equation and solve for x ,
[tex]\begin{gathered} 3x+2y=14 \\ \rightarrow3x+2(1)=14 \\ \rightarrow3x+2=14\rightarrow3x=12\rightarrow x=\frac{12}{3} \\ \\ \Rightarrow x=4 \end{gathered}[/tex]Therefore, we can conclude that the solution to our system is:
[tex]\begin{gathered} x=4 \\ y=1 \end{gathered}[/tex]