Question

4m+6n=54; 3m+2n=28
solve them

Answer 1

Given two equations i.e 4m+ 6n=54 (1)
3m+2n=28. (2)
Multiply eq 1 by 3 & eq 2by 4 we get,
12m+ 18n= 54×3. (3)
12m + 8n=28×4. (4)
Subtracting eq 4 from 3 we get,
12 m+ 18 n=162.
12m+ 8n= 112
(-) (-) (-)
=0+10n = 50
= 10n=50-0 =50
=n=50÷10=5
putting the value of n in 1 we get,
4m+ 6(5)= 54
4m+30=54
=4m= 54-30=14
=m=14÷4=7/2
Hope it helps you,dear.

Answer 2

$\underline{\bold{Answer:-}}$

Solving by Substitution method:-

$4m + 6n = 54 - - (1)\\ \\ 3m + 2n = 28 - - (2)$


From eq'n ( 1 )

$4m + 6n = 54 \\ \\ = > 6n = 54 - 4m \\ \\ = > n = ( \frac{54 - 4m}{6} ) - - (3)$

Substitute eq'n ( 3 ) in eq'n ( 2 )

$3m + 2n = 28 \\ \\ = > 3m + 2( \frac{54 - 4m}{6} ) = 28 \\ \\ = > 3m + ( \frac{108 - 8m}{6} ) = 28 \\ \\ = > 18m + 108 - 8m = 168 \\ \\ = > 10m = 60 \\ \\ = > m = \frac{60}{10} \\ \\ = > m = 6$

From eq'n ( 3 )

$n = ( \frac{54 - 4m}{6} ) \\ \\ = > n = ( \frac{54 - 4(6)}{6} )\\ \\ = > n = \frac{54 - 24}{6} \\ \\ = > n = \frac{30}{6} \\ \\ = > n = 5$

Therefore, the value of m = 6 and n = 5.

$\textbf{Hope it helps!}$