Question

Which of the following is the solution of the equation \[4x + 3y = 16\] ?
A. (1,3)
B. (1,4)
C. (2,3)
D. (2,5)

Answer 1

Hint: We have given only one 2 variable equation and we have to find the solution. We know different types of methods like elimination method, substation method etc. when we have 2 equations. But here, we will check each option in the equation whether it satisfies the equation or not.

Complete step by step answer:
We have given the equation \[4x + 3y = 16\]. We will put each option in the equation and try to satisfy the equation. If the L.H.S. is equal to RHS then the option is the solution of the equation. We will put each option in equation \[4x + 3y = 16\]. We will put (1,3) in \[4x + 3y = 16\].
\[ \Rightarrow 4x + 3y = 16\]
\[ \Rightarrow 4 \times 1 + 3 \times 3 = 16\]
\[ \Rightarrow 4 + 9 = 16\]
\[ \Rightarrow 13 \ne 16\]
So, LHS is not equal to RHS. We will now put (1,4) in \[4x + 3y = 16\]
\[ \Rightarrow 4x + 3y = 16\]
\[ \Rightarrow 4 \times 1 + 3 \times 4 = 16\]
\[ \Rightarrow 4 + 12 = 16\]
\[ \Rightarrow 16 = 16\]

So, LHS is equal to RHS. We will put (2,3) in \[4x + 3y = 16\]
\[ \Rightarrow 4x + 3y = 16\]
\[ \Rightarrow 4 \times 2 + 3 \times 3 = 16\]
\[ \Rightarrow 8 + 9 = 16\]
\[ \Rightarrow 17 \ne 16\]
So, LHS is not equal to RHS. We will also check (2,5) in \[4x + 3y = 16\]
\[ \Rightarrow 4x + 3y = 16\]
\[ \Rightarrow 4 \times 2 + 3 \times 5 = 16\]
\[ \Rightarrow 8 + 15 = 16\]
\[ \therefore 23 \ne 16\]
So, LHS is not equals to RHS

Hence, the option B is the correct answer.

Note: We should be familiar with different methods for solving two variable equations. We can use elimination methods, substitution methods for solving the equation.For example, in the substation method we first express one variable in terms of another variable then put the value of that variable in another equation then solve the equation.