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mylen [45]
3 years ago
14

What is the solution of this system of linear equations? 3y = 3 y equals StartFraction 3 over 2 EndFraction x plus 6.X + 6 y – S

tartFraction one-half EndFraction y minus StartFraction 1 over 4 EndFraction x equals 3.X = 3 (3, 6) (2, 1) no solution infinite number of solutions
Mathematics
2 answers:
Gennadij [26K]3 years ago
6 0

The solution of the given system of equation is \boxed{\bf (0,6)}.

Further explanation:

The given system of equations is as follows:

\boxed{\begin{aligned}y&=\dfrac{3}{2}x+6\\ \dfrac{y}{2}-\dfrac{x}{4}&=3\end{aligned}}

Label the above equations as follows:

y&=\dfrac{3}{2}x+6                                      ......(1)

\dfrac{y}{2}-\dfrac{x}{4}&=3                         ......(2)

To obtain the solution of the given system of equation use the substitution method.

Substitute the expression y&=\frac{3}{2}x+6 in equation (2) to obtain the value of x.

\begin{aligned}\dfrac{1}{2}\left(\dfrac{3}{2}x+6\right)-\dfrac{x}{4}&=3\\\dfrac{3}{4}x+3-\dfrac{x}{4}&=3\\\dfrac{3x-x}{4}+3-3&=0\\\dfrac{2x}{4}&=0\\x&=0\end{aligned}

Therefore, the value of x is 0.

Substitute 0 for x in equation (1) to obtain the value of y.

\begin{aligned}y&=\left(\dfrac{3}{2}\cdot 0\right)+6\\&=6\end{aligned}

Therefore, the value of y is 6.

From the above calculation it is concluded that the solution of the given system of equation is (0,6).

Thus, the solution of the given system of equation is \boxed{\bf (0,6)}.

Learn more:

1. A problem on composite function brainly.com/question/2723982  

2. A problem to find radius and center of circle brainly.com/question/9510228  

3. A problem to determine intercepts of a line brainly.com/question/1332667  

Answer details:

Grade: High school

Subject: Mathematics

Chapter: Linear equation

Keywords: Equation, linear equation, degree 1, higest power 1, system of linear equation, solution set, solution, mathematics, substitution method, consistent system , inconsistent system.

Scrat [10]3 years ago
3 0

Answer:

(0,6)

Step-by-step explanation:

The given system of equations is

y =  \frac{3}{2}x + 6

and

\frac{1}{2}y -  \frac{1}{4}x = 3

We substitute the first equation into the second equation to get:

\frac{1}{2} ( \frac{3}{2}x + 6) -  \frac{1}{4}x = 3

We expand to get:

\frac{3}{4} x + 3 -  \frac{1}{4}x = 3

We group similar terms to get:

\frac{3}{4}x -  \frac{1}{4}x = 3 - 3

\frac{1}{2}x = 0

x = 0

Put x=0 in to the first equation to get:

y = 6

Therefore the solution is (0,6)

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The product is the difference of squares is $$\left(11-b\right)\left(11+b\right)=121-{{b}^2}$$

Step-by-step explanation:

Explanation

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  • We have to multiply the given expression.
  • Square the first term 11. Square the last term b.

$$\begin{aligned}&(11-b)(11+b)=(11)^{2}-(b)^{2} \\&(11-b)(11+b)=121-b^{2}\end{aligned}$$

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3 years ago
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3 years ago
Give an example of how you use positive and negative numbers, and explain the meaning of 0 in your example.
EastWind [94]

Answer:

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How I would use them in every day life:

I would use positive numbers to represent... maybe something I'm trying to get at a store. Ex: 4 bananas

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2 years ago
Drag the tiles to the boxes to form correct pairs.<br> Match the pairs of equivalent expressions.
Rashid [163]

Answer:

The following pairs/results are matched:

  • 5\left(2t+1\right)+\left(-7t+28\right) = 3t+33
  • 3\left(3t-4\right)-\left(2t+10\right) = 7t-22
  • \left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right) = \frac{16t}{3}-\frac{23}{5}
  • \left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right) = -\frac{11}{4}t+36

Step-by-step explanation:

Lets solve all the expressions to match the results.

  • 5\left(2t+1\right)+\left(-7t+28\right)

<em>Solving the expression</em>

5\left(2t+1\right)+\left(-7t+28\right)

\mathrm{Remove\:parentheses}:\quad \left(-a\right)=-a

5\left(2t+1\right)-7t+28

10t+5-7t+28

3t+33

Therefore, 5\left(2t+1\right)+\left(-7t+28\right) = 3t+33

  • 3\left(3t-4\right)-\left(2t+10\right)

<em>Solving the expression</em>

3\left(3t-4\right)-\left(2t+10\right)

9t-12-\left(2t+10\right)

9t-12-2t-10

7t-22

Therefore, 3\left(3t-4\right)-\left(2t+10\right) = 7t-22

  • \left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)

<em>Solving the expression</em>

\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)

\mathrm{Remove\:parentheses}:\quad \left(a\right)=a

4t-\frac{8}{5}-\left(3-\frac{4}{3}t\right)

4t-\frac{8}{5}-\left(-\frac{4t}{3}+3\right)

4t-\frac{8}{5}-3+\frac{4t}{3}

As

-3-\frac{8}{5}:\quad -\frac{23}{5}    and  \frac{4t}{3}+4t:\quad \frac{16t}{3}

So,

4t-\frac{8}{5}-3+\frac{4t}{3} will become \frac{16t}{3}-\frac{23}{5}

Therefore, \left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right) = \frac{16t}{3}-\frac{23}{5}

  • \left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)

<em>Solving the expression</em>

\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)

\mathrm{Remove\:parentheses}:\quad \left(a\right)=a

-\frac{9}{2}t+3+\frac{7}{4}t+33

\mathrm{Group\:like\:terms}

\frac{9}{2}t+\frac{7}{4}t+3+33

\mathrm{Add\:similar\:elements:}\:-\frac{9}{2}t+\frac{7}{4}t=-\frac{11}{4}t

-\frac{11}{4}t+3+33

-\frac{11}{4}t+36

Therefore, \left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right) = -\frac{11}{4}t+36

Thus, the following pairs/results are matched:

  • 5\left(2t+1\right)+\left(-7t+28\right) = 3t+33
  • 3\left(3t-4\right)-\left(2t+10\right) = 7t-22
  • \left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right) = \frac{16t}{3}-\frac{23}{5}
  • \left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right) = -\frac{11}{4}t+36

Keywords: algebraic expression

Learn more about algebraic expression from brainly.com/question/11336599

#learnwithBrainly

5 0
3 years ago
Read 2 more answers
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