Question

How do you solve $3y=-\left( \dfrac{1}{2} \right)x+2$ and $y=-x+9$ using substitution?

Answer 1

Hint: To start with the problem, we will deal with it in a substitution method. From the first equation, we will start with finding the value of x. Putting the value of x in the second equation, we are getting the value of y. Thus we are finding the x and y using the substitution method.

Complete step-by-step solution:
According to the question, we are given two equations, $3y=-\left( \dfrac{1}{2} \right)x+2$ and $y=-x+9$. And we are to solve this problem using a substitution method.
In this problem, we will try to substitute x to solve the equations.
So, now from the first equation, $3y=-\left( \dfrac{1}{2} \right)x+2$
Multiplying both sides with 2 we are getting,
$\Rightarrow 6y=-x+4$
Now, changing sides, $x=4-6y$
Next, we will try to work with the second equation.
So, we have our second equation as, $y=-x+9$
Again, we also have, $x=4-6y$.
Putting the value of x in the second equation, we are getting,
$\Rightarrow y=-\left( 4-6y \right)+9$
Multiplying the negative sign inside, we get, $y=-4+6y+9$
Now, we will try to bring the terms with y in one side and the numbers on the other side of the equation.
Thus, we get, $y-6y=-4+9$
After further simplification, here we have, - 5y = 5
Dividing both sides with -5, we are getting, y = - 1.
Now, putting the value of y in the second equation to find x.
 $\Rightarrow -1=-x+9$
Further simplifying, x = 10,
So, the solution of the problem would be, x = 10, y = -1.

Note: The method of solving "by substitution" works by solving one of the equations (you choose which one) for one of the variables (you choose which one), and then plugging this back into the other equation, "substituting" for the chosen variable and solving for the other. Then you back-solve for the first variable. Here is how it works. Here we can also use the elimination method where we make a variable of same coefficient and of opposite sign and add both the equations for eliminating this variable.