Question:
Determine the value of m and n so that the following system of linear equation have an infinite number of solutions.
[tex](2m-1)x+3y-5=0[/tex]
and
[tex]3x+(n-1)y-2=0[/tex]
Answer:
[tex]m=\frac{17}{4}[/tex]
[tex]n = \frac{11}{5}[/tex]
Step-by-step explanation:
Given
[tex]3x+(n-1)y-2=0[/tex]
[tex](2m-1)x+3y-5=0[/tex]
Rewrite both equations:
[tex]3x + (n - 1)y = 2[/tex]
[tex](2m - 1)x + 3y = 5[/tex]
For the expression to have a solution, the following condition must exist:
[tex]\frac{3x}{(2m - 1)x} = \frac{(n-1)y}{3y}= \frac{2}{5}[/tex]
Split to 2
[tex]\frac{3x}{(2m - 1)x} = \frac{2}{5}[/tex]
[tex]\frac{(n-1)y}{3y}= \frac{2}{5}[/tex]
Cross Multiply
[tex]5 * 3x = 2 * (2m - 1)x[/tex] -- (1)
[tex]5 * (n -1)y = 2 * 3y[/tex] -- (2)
Solving (1)
[tex]5 * 3x = 2 * (2m - 1)x[/tex]
Divide both sides by x
[tex]5 * 3 = 2 * (2m - 1)[/tex]
Open bracket
[tex]15 = 4m - 2[/tex]
Collect Like Terms
[tex]4m = 15+2[/tex]
[tex]4m= 17[/tex]
Make m the subject
[tex]m=\frac{17}{4}[/tex]
Solving (2)
[tex]5 * (n -1)y = 2 * 3y[/tex]
Divide both sides by 7
[tex]5 * (n - 1) = 2 * 3[/tex]
Open bracket
[tex]5n - 5 = 6[/tex]
Collect Like Terms
[tex]5n = 5 + 6[/tex]
[tex]5n = 11[/tex]
Make n the subject
[tex]n = \frac{11}{5}[/tex]
