Sure! Let's solve the system of simultaneous linear equations step by step to find the value of [tex]\( x + y \)[/tex].
We are given the equations:
[tex]\[ 7x + 2y = 9 \quad (1) \][/tex]
[tex]\[ 2x + 7y = 27 \quad (2) \][/tex]
To eliminate one of the variables, we can use the method of elimination. We will multiply both equations by suitable numbers so that the coefficients of one of the variables become the same.
Let's start by eliminating [tex]\( y \)[/tex]. To do this, multiply equation (1) by 7 and equation (2) by 2:
[tex]\[ 49x + 14y = 63 \quad (3) \][/tex]
[tex]\[ 4x + 14y = 54 \quad (4) \][/tex]
Next, subtract equation (4) from equation (3):
[tex]\[ (49x + 14y) - (4x + 14y) = 63 - 54 \][/tex]
[tex]\[ 45x = 9 \][/tex]
[tex]\[ x = \frac{9}{45} \][/tex]
[tex]\[ x = \frac{1}{5} \][/tex]
Now that we have the value of [tex]\( x \)[/tex], substitute [tex]\( x = \frac{1}{5} \)[/tex] back into one of the original equations to solve for [tex]\( y \)[/tex]. Let's use equation (1):
[tex]\[ 7 \left( \frac{1}{5} \right) + 2y = 9 \][/tex]
[tex]\[ \frac{7}{5} + 2y = 9 \][/tex]
[tex]\[ 2y = 9 - \frac{7}{5} \][/tex]
[tex]\[ 2y = \frac{45}{5} - \frac{7}{5} \][/tex]
[tex]\[ 2y = \frac{38}{5} \][/tex]
[tex]\[ y = \frac{38}{10} \][/tex]
[tex]\[ y = \frac{19}{5} \][/tex]
Now, we have the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ x = \frac{1}{5} \][/tex]
[tex]\[ y = \frac{19}{5} \][/tex]
We need to find the value of [tex]\( x + y \)[/tex]:
[tex]\[ x + y = \frac{1}{5} + \frac{19}{5} \][/tex]
[tex]\[ x + y = \frac{1 + 19}{5} \][/tex]
[tex]\[ x + y = \frac{20}{5} \][/tex]
[tex]\[ x + y = 4 \][/tex]
Thus, the value of [tex]\( x + y \)[/tex] is [tex]\( 4 \)[/tex], and the correct answer is:
[tex]\[ \boxed{4} \][/tex]
