onsider the pair of simultaneous equations: $\newline$ $\begin{array}{l} 2 x+y=5 \ldots \\ 5 x+y=11 \ldots \end{array}$ $\newline$ Solve the equations by first subtracting equation [ $2$ ] from equation [ $1$ ], i.e. [ $1$ ] - [ $2$ ]. Now solve the equations by first subtracting equation [ $1$ ] from equation [ $2$ ], i.e. [ $2$ ] - [ $1$ ]. Which method a or $b$ is preferable and why?

Full solution

Q. onsider the pair of simultaneous equations:

\newline

\begin{array}{l} 2 x+y=5 \ldots \\ 5 x+y=11 \ldots \end{array}

\newline

Solve the equations by first subtracting equation [

2

] from equation [

1

], i.e. [

1

] - [

2

]. Now solve the equations by first subtracting equation [

1

] from equation [

2

], i.e. [

2

] - [

1

]. Which method a or

b

is preferable and why?

Write Equations: First, let's write down the equations again: $\newline$ Equation [ $1$ ]: $2x + y = 5$ $\newline$ Equation [ $2$ ]: $5x + y = 11$
Subtract Equations: Now, subtract equation [ $2$ ] from equation [ $1$ ]: $\newline$ $(2x + y) - (5x + y) = 5 - 11$ $\newline$ This simplifies to: $\newline$ $-3x = -6$
Solve for x: Divide both sides by $-3$ to solve for x: $\newline$ $x = \frac{-6}{-3}$ $\newline$ $x = 2$
Plug in $x$ for $y$ : Now, plug $x = 2$ into equation [ $1$ ] to solve for $y$ : $2(2) + y = 5$ $4 + y = 5$ $y = 5 - 4$ $y = 1$
Try Other Method: Now let's try the other method, subtracting equation [ $1$ ] from equation [ $2$ ]: $\newline$ $(5x + y) - (2x + y) = 11 - 5$ $\newline$ This simplifies to: $\newline$ $3x = 6$
Solve for x: Divide both sides by $3$ to solve for x: $\newline$ $x = \frac{6}{3}$ $\newline$ $x = 2$
Plug in $x$ for $y$ : Now, plug $x = 2$ into equation [ $2$ ] to solve for $y$ : $5(2) + y = 11$ $10 + y = 11$ $y = 11 - 10$ $y = 1$
Compare Methods: Both methods give us the same solution, $x = 2$ and $y = 1$ . However, method b (subtracting equation [ $1$ ] from equation [ $2$ ]) is preferable because the coefficients of $x$ are positive, which can be easier to work with.