We have,<\/p>\n
2x + 3y = 11\u00a0 \u00a0 \u00a0 \u00a0 ….(1)<\/p>\n
and 2x – 4y = -24\u00a0 \u00a0 \u00a0 \u00a0 ……(2)<\/p>\n
From equation (1), we have 2x = 11 – 3y<\/p>\n
Substituting the value of 2x = 11 – 3y in (2), we get<\/p>\n
11 – 3y – 4y = -24<\/p>\n
-7y = -24 – 11<\/p>\n
-7y = -35\u00a0 \u00a0 \\(\\implies\\)\u00a0 \u00a0y = 5<\/p>\n
Putting y = 5 in equation (1), we get<\/p>\n
2x + 3 \\(\\times\\) = 11<\/p>\n
2x = 11 – 15<\/p>\n
\\(\\implies\\)\u00a0 x = -\\(4\\over 2\\) = -2<\/p>\n
Hence, x = -2\u00a0 and y = 5<\/p>\n
Now put x = – 2 and y = 5 in y = mx + 3, we get<\/p>\n
5 = m(-2) + 3\u00a0 \u00a0 \u00a0 \\(\\implies\\)\u00a0 \u00a0-2m = 5 – 3<\/p>\n