Calculate the solution to these simultaneous equations using the substitution method: (1)(1)4𝑚+3𝑛4m+3n==3232 (2)(2)𝑚m==𝑛+1n+1

Sure, let's solve these equations step by step:

1. We have two equations here:

   Equation 1: 4m + 3n = 32 Equation 2: m = n + 1

2. We can substitute Equation 2 into Equation 1. Wherever we see 'm' in Equation 1, we can replace it with 'n+1' (from Equation 2).

   So, Equation 1 becomes: 4(n + 1) + 3n = 32

3. Simplify the equation:

   4n + 4 + 3n = 32 7n + 4 = 32

4. Subtract 4 from both sides to isolate the term with 'n':

   7n = 32 - 4 7n = 28

5. Divide both sides by 7 to solve for 'n':

   n = 28 / 7 n = 4

6. Now that we have the value for 'n', we can substitute it back into Equation 2 to find the value for 'm':

   m = n + 1 m = 4 + 1 m = 5

So, the solution to the simultaneous equations is m = 5 and n = 4.