Question

How do you solve this system of equations: $4s-3t=8$ and $t=-2s-1?$

Answer 1

Hint: We will use elimination method to solve this system of equations. We will check if it is possible to eliminate any of the unknown variables by the operations, addition or subtraction. If it cannot be done directly, we will multiply the equations with suitable scalars so that the coefficients of only one variable become the same in both the equations. Then we will eliminate the variables.

Complete step by step answer:
Let us consider the given system of equations, $4s-3t=8$ and $t=-2s-1.$
We are asked to solve the system of equations. We will use the elimination method to solve this system of equations. We will eliminate one of the variables to find the value of the other variable.
Consider the equation $4s-3t=8.......\left( 1 \right)$
Now consider the equation $t=-2s-1.......\left( 2 \right)$
We can see that the equation $\left( 2 \right)$ is not in the standard form. First, let us write it in the standard form by transposing $2s$from the RHS to the LHS.
We will get $2s+t=-1.......\left( 3 \right)$
By inspection, we can see that the elimination of the variable $s$ can be done using the addition of the equation $\left( 1 \right)$ and a scalar multiple of the equation $\left( 3 \right).$
Let us multiply the equation$\left( 3 \right)$ with $-2$ to get $-4s-2t=2.......\left( 4 \right).$
When we add the equations $\left( 1 \right)$ and $\left( 4 \right),$ we can eliminate $s.$
So, we will get $4s-3t-4s-2t=8+2=10.$
After the operation is done, we will get $-3t-2t=-5t=10.$
Let us transpose $-5$ to get $t=\dfrac{10}{-5}=-2.$
When we apply the value of $t$ in the equation $\left( 1 \right),$ we will get $4s-3\times \left( -2 \right)=8.$
That is, $4s+6=8.$
Let us transpose $6$ to get $4s=8-6=2.$
Now we will transpose $4$ to get $s=\dfrac{2}{4}=\dfrac{1}{2}.$

Hence the solution is $s=\dfrac{1}{2}$ and $t=-2.$

Note: We know that we will get the same values for $s$ and $t$ if we multiply the equation $\left( 3 \right)$ with $2$ and subtract it from the equation $\left( 1 \right).$ We will get $4s-3t-\left( 4s+2t \right)=4s-3t-4s-2t=-5t=8+2=10.$ Now, we know that this is same as the equation we have obtained.