Math, asked by lorrdjames15, 9 months ago

Show and explain how replacing one equation by the sum of that equation and a multiple of the other produces a system with the same solutions as the one shown.

8x + 7y = 39

4x – 14y = –68

Answers

Answered by makaylahatkinson
61

Answer:

Sample Response: Using the linear combination method, you can multiply the first equation by 2 and add the equations to get 20x = 10. Dividing both sides by 20, x = 1/2. To solve for y, substitute 1/2 for x in the equation 8x + 7y = 39 to get 4 + 7y = 39. Solving this equation, y = 5. Checking this in the other equation, 4(1/2) – 14(5) = –68 results in 2 – 70 = –68 or –68 = –68. The solution of the system shown is (1/2, 5). The system 8x + 7y = 39 and 20x = 10 is formed by replacing 4x –14y = –68 by a sum of it and a multiple of 8x + 7y = 39. Since 20(1/2) = 10, the system 8x + 7y = 39 and 20x = 10 also has a solution of (1/2, 5).

Step-by-step explanation:

Here's the sample response it gives you on Edge to give you a basic Idea of what it's looking for. =)

The other guy had a good answer and it's correct as well sooo

Mark them as Brainliest or whatever.

Answered by masura8080
20

Answer:

We will get the value of x=\frac{1}{2} and y=5.

Step-by-step explanation:

  • We have to evaluate the given data.

         Given data:- 8x+7y=39, 4x-14y=-68.

         To find:- Value of x and y.

         Solution:-

  • First, let's multiply the first equation by two on both sides:

         [tex]2\times( 8 x+7 y=39)\\ =>16x+14y=78[/tex]

  • Now the equation is

          [tex]\begin{array}{l} 16 x+14 y=78 \\ 4 x-14 y=-68 \end{array}[/tex]

  • After adding this up in the column

      [tex]\begin{array}{l} (16 x+4 x)+(14 y-14 y)=78-68 \\ \Rightarrow20 x=10 \\ \Rightarrow x=\frac{10}{20} \\ \Rightarrow x=\frac{1}{2}\\ \end{array}[/tex]

  • y can be calculated by replacing the x

         [tex]\begin{array}{l} 8 x+7 y=39 \\ \Rightarrow 8 \times \frac{1}{2} +7 y=39 \\ \Rightarrow 4+7 y=39 \\ \Rightarrow 7 y=39-4 \\ \Rightarrow 7 y=35 \\ \Rightarrow y=35 \div 7=5 \end{array}[/tex]

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