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Aleks04 [339]
1 year ago
11

How can I solve this linear equation x+y=7 3x+y=15​

Mathematics
1 answer:
bezimeni [28]1 year ago
6 0

Answer:

  (x, y) = (4, 3)

Step-by-step explanation:

The first step in solving any math problem is to read and understand what is given and what is asked.

<h3>Given</h3>

Two linear equations are given, each in standard form. The coefficients of y are the same in the two equations, and both coefficients are 1 in the first equation.

  • x +y = 7
  • 3x +y = 15

<h3>Find</h3>

You are asked to "solve" this system of equations. That means you want an (x, y) ordered pair that will satisfy both equations. On a graph, these are the coordinates of the point where the lines cross.

<h3>Solution</h3>

There are many ways to solve a system of linear equations. In this case, the y-coefficients being the same suggests that "elimination" or "linear combination" would be a good choice of solution method.

We note that the coefficients in the second equation are generally larger than those in the first equation. Subtracting the first equation from the second equation will generally leave positive coefficients, making the arithmetic less error-prone.

  (3x +y) -(x +y) = (15) -(7) . . . . . . subtract the first equation from the second

  2x = 8 . . . . . . simplify

  x = 4 . . . . . . divide by 2

Using this value in the first equation, we can find y:

  4 +y = 7

  y = 3 . . . . . . subtract 4

The solution to this pair of equations is (x, y) = (4, 3).

__

<em>Additional comment</em>

A graphical solution is attached.

Either equation can be solved for y, and that expression substituted into the other. Solving the first equation for y gives ...

  y = 7 -x

Substituting that into the second equation, we have ...

  3x +(7 -x) = 15

  2x = 8 . . . . . . . subtract 7

  x = 4 . . . . . . divide by 2

  y = 7-4 = 3

(x, y) = (4, 3) is the solution.

We have shown 3 methods of solving the given system of equations. There are other methods that can be used, too, including matrix methods and methods that make use of arrays of the coefficients.

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