What is 3x+4y=1,(2k-1)+(k-1)y=2k+1 ?

The solution to 3x+4y=1,(2k-1)+(k-1)y=2k+1 is

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3x+4y=1,(2k-1)+(k-1)y=2k+1

3 x + 4 y = 1, (2 k - 1) + (k - 1) y = 2 k + 1

(x = \frac{k - 9}{3 ( k - 1 )}, y = \frac{2}{k - 1})

Solution steps

[3 x + 4 y = 1 (2 k - 1) + (k - 1) y = 2 k + 1]

Solve

(2 k - 1) + (k - 1) y = 2 k + 1 : y = \frac{2}{k - 1}; k \neq = 1

Plug the solutions

y = \frac{2}{k - 1}

into

3 x + 4 y = 1

For

3 x + 4 y = 1

, subsitute

y

with

\frac{2}{k - 1} : x = \frac{k - 9}{3 ( k - 1 )}

Verify solutions by plugging them into the original equations

Therefore, the final solution for

3 x + 4 y = 1, (2 k - 1) + (k - 1) y = 2 k + 1

(x = \frac{k - 9}{3 ( k - 1 )}, y = \frac{2}{k - 1})

m + 2 n = 4, 3 n - m = 16

x + 4 y = 3 x - 2, y = 5

0.5 x = 2, x = 0

2 x + 3 y = 4, 4 x - 3 y = 8

3 x + 4 y = 10, x - y = 1

What is 3x+4y=1,(2k-1)+(k-1)y=2k+1 ?
The solution to 3x+4y=1,(2k-1)+(k-1)y=2k+1 is