Mathematics » Equations and Inequalities » Simultaneous Equations Continued

Solving by Elimination II

Solving by Elimination

  • Make one of the variables the subject of both equations.

  • Equate the two equations; by doing this we reduce the number of equations and the number of variables by one.

  • We now have one equation with one unknown variable which can be solved.

  • Use the solution to substitute back into either original equation, to find the corresponding value of the other unknown variable.

Example

Question

Solve for \(x\) and \(y\): \begin{align*} y &= x^2 – 6x \qquad \ldots (1) \\ y + \cfrac{1}{2}x – 3 &= 0 \qquad \ldots (2) \end{align*}

Make \(y\) the subject of the second equation

\begin{align*} y + \cfrac{1}{2}x – 3 &= 0 \\ y &= -\cfrac{1}{2}x + 3 \end{align*}

Equate the two equations and solve for \(x\)

\begin{align*} x^2 – 6x &= -\cfrac{1}{2}x + 3 \\ x^2 – 6x + \cfrac{1}{2}x – 3 &= 0\\ 2x^2 – 12x + x – 6 &= 0\\ 2x^2 -11x – 6 &= 0\\ (2x +1)(x-6) &= 0\\ \text{Therefore } x = -\cfrac{1}{2}&\text{ or } x = 6 \end{align*}

Substitute the values for \(x\) back into the second equation to calculate the corresponding \(y\)-values

If \(x = -\dfrac{1}{2}\): \begin{align*} y &= -\cfrac{1}{2} ( -\cfrac{1}{2} ) + 3 \\ \therefore y &= 3\cfrac{1}{4} \end{align*} This gives the point \((-\dfrac{1}{2};3\dfrac{1}{4})\).

If \(x = 6\): \begin{align*} y &= -\cfrac{1}{2}(6) + 3 \\ &= -3 + 3 \\ \therefore y &= 0 \end{align*} This gives the point \((6;0)\).

Check that the two points satisfy both original equations

Write the final answer

The solution is \(x = -\dfrac{1}{2} \text{ and } y = 3\dfrac{1}{4}\) or \(x = 6 \text{ and } y = 0\). These are the coordinate pairs for the points of intersection as shown below.

52528403900da2fe061055f5d6d10b18.png

Example

Question

Solve for \(x\) and \(y\): \begin{align*} y &= \dfrac{5}{x – 2} \qquad \ldots (1)\\ y + 1 &= 2x \qquad \ldots (2) \end{align*}

Make \(y\) the subject of the second equation

\begin{align*} y + 1 &= 2x \\ y &= 2x – 1 \end{align*}

Equate the two equations and solve for \(x\)

\begin{align*} 2x – 1 &= \cfrac{5}{x – 2} \\ (2x – 1)(x – 2) &= 5 \\ 2x^2 – 5x + 2 &= 5 \\ 2x^2 – 5x – 3 &= 0 \\ (2x + 1)(x – 3) &= 0 \\ \text{Therefore } x = -\cfrac{1}{2}&\text{ or } x = 3 \end{align*}

Substitute the values for \(x\) back into the second equation to calculate the corresponding \(y\)-values

If \(x -\dfrac{1}{2}\): \begin{align*} y &= 2(-\cfrac{1}{2}) – 1 \\ \therefore y &= -2 \end{align*} This gives the point \((-\dfrac{1}{2};-2)\).

If \(x = 3\): \begin{align*} y &= 2(3) -1 \\ &= 5 \end{align*} This gives the point \((3;5)\).

Check that the two points satisfy both original equations

Write the final answer

The solution is \(x = -\dfrac{1}{2} \text{ and } y = -2\) or \(x = 3 \text{ and } y = 5\). These are the coordinate pairs for the points of intersection as shown below.

68ed42298fce291b00d3fa2af26547c9.png

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