Elimination by equating coefficients

For example, in the following system of equations, the variable $x$ is multiplied by the factor 3 in each equation:

One can then solve for $x$ and equate the other side in each case. Thus, one eliminates one variable (here: $x$) and can determine the other variable (here: $y$) by inserting it into one of the equations.

Example

The procedure will now be demonstrated with the following system of equations with $2$ equations and $2$ variables:

• $\text{𝐒𝐨𝐥𝐯𝐞 𝐛𝐨𝐭𝐡 𝐞𝐪𝐮𝐚𝐭𝐢𝐨𝐧𝐬 𝐟𝐨𝐫 𝐨𝐧𝐞 𝐯𝐚𝐫𝐢𝐚𝐛𝐥𝐞}$

First, both sides are solved for one variable. In this case, for example, it will be solved for $a$.

Since now at ${𝖨}^{\prime }$ and $𝖨{𝖨}^{\prime }$ the left-hand sides are both equal, the right-hand sides must also be equal, therefore $\mathrm{𝟧}-{\displaystyle \frac{\mathrm{𝟣}}{\mathrm{𝟤}}}𝖻=\mathrm{𝟥}+{\displaystyle \frac{\mathrm{𝟣}}{\mathrm{𝟤}}}𝖻$.

This step is called $\text{"𝐞𝐪𝐮𝐚𝐭𝐢𝐧𝐠"}$.

• $\text{𝐄𝐪𝐮𝐚𝐭𝐢𝐧𝐠 }\text{𝐈}\text{’}\text{𝐚}\text{𝐧}\text{𝐝}\text{𝐈}\text{𝐈}\text{’}$

$\mathrm{𝟧}-{\displaystyle \frac{\mathrm{𝟣}}{\mathrm{𝟤}}}𝖻=\mathrm{𝟥}+{\displaystyle \frac{\mathrm{𝟣}}{\mathrm{𝟤}}}𝖻$

• $\text{𝐒𝐨𝐥𝐯𝐞 𝐞𝐪𝐮𝐚𝐭𝐢𝐨𝐧}$

This new equation has $\text{𝐨𝐧𝐥𝐲 𝐨𝐧𝐞 𝐯𝐚𝐫𝐢𝐚𝐛𝐥𝐞}$ and can therefore be solved as usual.

This solution can now be substituted into one of the upper equations to calculate the value of the second variable.

• $\text{𝐒𝐮𝐛𝐬𝐭𝐢𝐭𝐮𝐭𝐢𝐨𝐧 𝐢𝐧𝐭𝐨 𝐞𝐪𝐮𝐚𝐭𝐢𝐨𝐧 }\text{𝐈}\text{’}\text{𝐨}\text{𝐫}\text{𝐈}\text{𝐈}\text{’}$

It does not matter which equation you use! To check the result, you can also insert it into both equations and check if the same value comes out.

Substitution of b into $𝖨{𝖨}^{\prime }$

$𝖺=\mathrm{𝟥}+{\displaystyle \frac{\mathrm{𝟣}}{\mathrm{𝟤}}}\cdot \mathrm{𝟤}=\mathrm{𝟥}+\mathrm{𝟣}=\mathrm{𝟦}$

This yields the solution set:

$𝖫=\{(\mathrm{𝟦};\mathrm{𝟤})\}$

• $\text{𝐏𝐫𝐨𝐨𝐟 (𝐜𝐚𝐧 𝐚𝐥𝐬𝐨 𝐛𝐞 𝐨𝐦𝐦𝐢𝐭𝐞𝐝) }$

To check the solution, substitute it into the original equation and check that they are satisfied.