For the system of linear equations x/3-y/5=4 and 3x/4+2y/3=6,how many solutions exist?



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giorgiana1976's profile pic

Posted on (Answer #1)

We'll find the least common denominator for the first two fractions from the top equation.

5x - 3y = 60 (1)

We'll find the least common denominator for the next two fractions from the bottom equation.

9x + 8y = 72 (2)

We'll use elimination method to determine the solution of the system.

We'll isolate x from the 1st equation, to the left side.

x = (60+3y)/5

We'll replace x into the 2nd equation and we'll get an equation in y only:

9(60+3y)/5 + 8y = 72

540 + 27y + 40y = 360

67y = 360-540

67y = -180

y = -180/67

x = (60 - 540/67)/5

x = 3480/67*5

x = 696/67

The solution of the system is the pair of coordinates (696/67 ; -180/67).

samhouston's profile pic

Posted on (Answer #2)

A system of two linear equations has only 3 possible outcomes:

* no solution

* 1 solution

* infinite solutions

You can tell which of these is correct by graphing.  If the lines are parallel, then there is no solution.  If the lines intersect at one point, then there is one solution.  If the lines coincide, then there are infinite solutions.

 x/3-y/5=4 and 3x/4+2y/3=6

First, rewrite equation in slope-intercept form.

x/3 - y/5 = 4

y = 5(x/3 - 4)

3x/4 + 2y/3 = 6

y = 3/2(6 - 3x/4)

Now graph the two equations.


The lines intersect at exactly one point, therefore this system has one solution.

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