11.1 Systems of linear equations: two variables  (Page 4/20)

Both equations are already set equal to a constant. Notice that the coefficient of $x$ in the second equation, –1, is the opposite of the coefficient of $x$ in the first equation, 1. We can add the two equations to eliminate $x$ without needing to multiply by a constant.

Now that we have eliminated $x,$ we can solve the resulting equation for $y.$

Then, we substitute this value for $y$ into one of the original equations and solve for $x.$

The solution to this system is $\left(-\frac{7}{3},\frac{2}{3}\right).$

Check the solution in the first equation.

Adding these equations as presented will not eliminate a variable. However, we see that the first equation has $3x$ in it and the second equation has $x.$ So if we multiply the second equation by $\mathrm{-3},$ the x -terms will add to zero.

Now, let’s add them.

For the last step, we substitute $y=\mathrm{-4}$ into one of the original equations and solve for $x.$

Our solution is the ordered pair $\left(3,\mathrm{-4}\right).$ See [link] . Check the solution in the original second equation.

One equation has $2x$ and the other has $5x.$ The least common multiple is $10x$ so we will have to multiply both equations by a constant in order to eliminate one variable. Let’s eliminate $x$ by multiplying the first equation by $\mathrm{-5}$ and the second equation by $2.$

Then, we add the two equations together.

Substitute $y=\mathrm{-4}$ into the original first equation.

The solution is $\left(\mathrm{-2},\mathrm{-4}\right).$ Check it in the other equation.

See [link] .

First clear each equation of fractions by multiplying both sides of the equation by the least common denominator.

Now multiply the second equation by $\mathrm{-1}$ so that we can eliminate the x -variable.

Add the two equations to eliminate the x -variable and solve the resulting equation.

Substitute $y=7$ into the first equation.

The solution is $\left(\frac{11}{2},7\right).$ Check it in the other equation.