Question

How do you solve the system of linear equations \[ - x - 5y = 1\] and \[ - 3x + 7y = 25\] by combination?

Answer 1

Hint: We use the combination method to solve two linear equations given in the question. We find the value of x from the first equation in terms of y and substitute in the second equation which becomes an equation in y entirely. Solve for the value of y and substitute back the value of y to obtain the value of x.
* Combination method of solving systems of linear equations: We multiply the equation (or equations) with such constant values that make one of the values exactly equal in both linear equations. According to the signs of that value we add or subtract the equations.

Complete step-by-step answer:
We have two linear equations\[ - x - 5y = 1\] … (1)
And \[ - 3x + 7y = 25\] … (2)
We multiply equation (1) with constant 3
\[ \Rightarrow - 3x - 15y = 3\] … (3)
Now we subtract equation (3) from equation (2)
\[\begin{gathered}
   - 3x + 7y = 25 \\
  \underline { - 3x - 15y = 3} \\
  0x + 22y = 22 \\
\end{gathered} \]
Now we get the equation as \[22y = 22\]
Cancel same factors from both sides of the equation
\[ \Rightarrow y = 1\] … (4)
Substitute this value of y back in equation (1) to calculate value of x
\[ \Rightarrow - x - 5 \times 1 = 1\]
\[ \Rightarrow - x - 5 = 1\]
Shift all constant values to right hand side of the equation
\[ \Rightarrow - x = 1 + 5\]
\[ \Rightarrow - x = 6\]
Multiply both sides of the equation by -1
\[ \Rightarrow x = - 6\]
So, value of x is -6 and value of y is 1

\[\therefore \]Solution of the system of linear equations \[ - x - 5y = 1\] and \[ - 3x + 7y = 25\] is \[x = - 6;y = 1\]

Note:
Many students get confused between combination method and substituting method but they can remember the combination method by linking it to ‘linear combination’ which means one equation is linear combination of other i.e. we can convert one term of an equation with same variable equal to term of same variable of other equation. Also, always remember to change sign from positive to negative and vice-versa when subtracting equations.