Mathematics » Equations and Inequalities » Solving Simultaneous Equations

Solving by Elimination

Solving by Elimination

Example

Question

Solve the following system of equations:

\begin{align*} 3x + y & = 2 \qquad \ldots (1) \\ 6x – y & = 25 \quad \ldots (2) \end{align*}

Make the coefficients of one of the variables the same in both equations

The coefficients of \(y\) in the given equations are \(\text{1}\) and \(-\text{1}\). Eliminate the variable \(y\) by adding equation \((1)\) and equation \((2)\) together:

\[\begin{array}{cccc} & 3x + y & = & 2 \\ + & 6x – y & = & 25 \\ \hline & 9x + 0 & = & 27 \end{array}\]

Simplify and solve for \(x\)

\begin{align*} 9x & = 27 \\ \therefore x & = 3 \end{align*}

Substitute \(x\) back into either original equation and solve for \(y\)

\begin{align*} 3(3) + y & = 2 \\ y & = 2 – 9 \\ \therefore y & = -7 \end{align*}

Check that the solution \(x=3\) and \(y=-7\) satisfies both original equations

Write the final answer

\begin{align*} x & =3 \\ y & =-7 \end{align*}

Example

Question

Solve the following system of equations:

\begin{align*} 2a – 3b & = 5 \qquad \ldots (1) \\ 3a – 2b & = 20 \qquad \ldots(2) \end{align*}

Make the coefficients of one of the variables the same in both equations

By multiplying equation \((1)\) by \(\text{3}\) and equation \((2)\) by \(\text{2}\), both coefficients of \(a\) will be \(\text{6}\).

\[\begin{array}{cccc} & 6a – 9b & = & 15 \\ – & (6a – 4b & = & 40) \\ \hline & 0 – 5b & = & -25 \end{array}\]

(When subtracting two equations, be careful of the signs.)

Simplify and solve for \(b\)

\begin{align*} b & = \cfrac{-25}{-5} \\ \therefore b & = 5 \end{align*}

Substitute value of \(b\) back into either original equation and solve for \(a\)

\begin{align*} 2a – 3(5) & = 5 \\ 2a – 15 & = 5 \\ 2a & = 20 \\ \therefore a & = 10 \end{align*}

Check that the solution \(a=10\) and \(b=5\) satisfies both original equations

Write the final answer

\begin{align*} a & = 10 \\ b & = 5 \end{align*}

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