The given linear equations are :<\/p>\n
8x + 5y = 9\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ………..(1)<\/p>\n
3x + 2y = 4\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ………..(2)<\/p>\n
By Substitution Method :<\/strong><\/p>\n
Form equation (1),\u00a0 y = \\(9 – 8x\\over 5\\)<\/p>\n
Substitute the value of y in equation (2), we get<\/p>\n
3x + 2(\\(9 – 8x\\over 5\\)) = 4<\/p>\n
or\u00a0 \u00a015x + 18 – 16x = 20\u00a0 \u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0x = -2<\/p>\n
Putting x = – 2 in equation (1), we get<\/p>\n
8(-2) + 5y = 9\u00a0 \u00a0 \\(\\implies\\)\u00a0 \u00a05y = 9 + 16 = 25<\/p>\n
\\(\\implies\\)\u00a0 y = 5<\/p>\n
Hence,\u00a0 x = -2 and y = 5.<\/strong><\/p>\n
By Cross-Multiplication Method :<\/strong><\/p>\n
The given linear equation can be written as :<\/p>\n
8x + 5y – 9 = 0<\/p>\n
3x + 2y – 4 = 0<\/p>\n
So, we have<\/p>\n
\\(x\\over -20 + 18\\)\u00a0 =\u00a0 \\(y\\over -27 + 32\\)\u00a0 =\u00a0 \\(1\\over 16 – 15\\)<\/p>\n
\\(\\implies\\)\u00a0 \u00a0\\(x\\over -2\\)\u00a0 =\u00a0 \\(y\\over\u00a0 5\\)\u00a0 =\u00a0 \\(1\\over 1\\)<\/p>\n
\\(\\implies\\)\u00a0 \u00a0x = -2\u00a0 and y = 5<\/strong><\/p>\n\n\n
<\/p>\n\n\n\n
<\/p>\n","protected":false},"excerpt":{"rendered":"
Solution : The given linear equations are : 8x + 5y = 9\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ………..(1) 3x + 2y = 4\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ………..(2) By Substitution Method : Form equation (1),\u00a0 y = \\(9 – 8x\\over 5\\) Substitute the value of y in equation (2), we get …<\/p>\n