Students who successfully complete this primer will be able to:

Identify situations that involve solving two equations and two unknowns.

Use substitution to obtain one equation and one unknown.

Simplify the above result and solve for both of the unknowns.

Before working on this primer, you may want to review:

Solving a linear equation

Often in physics, we have a situation $($such as projectile motion or the sum of forces in two dimensions$)$ where we have two equations with the same two unknowns in each. The most common method is substitution, which generally follows these steps:

$1.$ Isolate one of the variables in one of the two equations. Let's say that we have

$2$A$$B$=02$A$$B$=0$

A$+2$B$=5$A$+2$B$=5$

$$We can isolate$$BB in $($a$)$ to give B$=2$AB$=2$A$.$

$2.$ Substitute this result into $($b$)$:

$$ b$.$A$+2(2$A$)=5$A$+2(2$A$)=5$

$3.$ Solve for the unknown that remains:

$$ b$.$ A$+4$A$=55$A$=5$A$=1$A$+4$A$=55$A$=5$A$=1$

$4.$ Substitute this result into either of the original two equations to solve for the other variable:

$$ a$.2(1)$B$=0$B$=22(1)$B$=0$B$=2$

Part A $-$ Isolating a Variable

Isolating a variable in two equations is easiest when one of them has a coefficient of $1.$ Let's say we have the two equations

$3$A$$B$=53$A$$B$=5$

$2$A$+3$B$=42$A$+3$B$=4$

and want to isolate one of the variables, such that it appears by itself on one side of the equation. Which of the following is an equation with one of the above variables isolated?

View Available Hint$($s$)$for Part A

$2$A$=3$B$42$A$=3$B$4$B$=53$AB$=53$AB$=3$A$5$B$=3$A$53$B$=2$A$43$B$=2$A$4$

Submit

Part B $-$ Substitution

Now that we have one of the variables from Part A isolated, and written in terms of the other variable, we can now substitute this into the other of the two original equations. Which of the following options represents this?

$3$A$(3$A$5)=43$A$(3$A$5)=42($B$+5)+3$B$=42($B$+5)+3$B$=42$A$+3(5)=42$A$+3(5)=42$A$+3(3$A$5)=42$A$+3(3$A$5)=4$

SubmitRequest$$Answer

Part C $-$ Solving for the Variables

We now have an algebraic expression with only one variable, which can be solved. Once we have that, we can plug it back into one of the original equations $($or the expression derived in Part A$)$ to solve for the other variable. When this is done with the system of two equations from Parts A and B$,$ what is the solution?

Enter AA$,$ then BB$,$ as two numbers, separated by a comma.

View Available Hint$($s$)$for Part C

Activate to select the appropriates template from the following choices. Operate up and down arrow for selection and press enter to choose the input value typeActivate to select the appropriates symbol from the following choices. Operate up and down arrow for selection and press enter to choose the input value type

AA$,$ BB $=$

nothing

Submit

Part D $-$ Isolating a Variable with a Coefficient

In some cases, neither of the two equations in the system will contain a variable with a coefficient of $1,$ so we must take a further step to isolate it$.$Let$\text{'}$s say we now have

$3$C$+4$D$=53$C$+4$D$=5$

$2$C$+5$D$=22$C$+5$D$=2$

None of these terms has a coefficient of $1.$ Instead, we'll pick the variable with the smallest coefficient and isolate it$.$ Move the term with the lowest coefficient so that it's alone on one side of its equation, then divide by the coefficient. Which of the following expressions would result from that process?

$$C$=152$DC$=152$D$$D$=5434$CD$=5434$C$$C$=5343$DC$=5343$D$$D$=2525$CD$=2525$C$$

SubmitRequest$$Answer

Part E $-$ Solving for Two Variables

Now that you have one of the two variables in Part D isolated, use substitution to solve for the two variables.$$You may want to review the Multiplication and Division of Fractions and Simplifying an Expression Primers.

Enter the answer as two numbers $($either fraction or decimal$),$ separated by a comma, with CC first.

View Available Hint$($s$)$for Part E

Activate to select the appropriates template from the following choices. Operate up and down arrow for selection and press enter to choose the input value typeActivate to select the appropriates symbol from the following choices. Operate up and down arrow for selection and press enter to choose the input value type

CC$,$ DD $=$

nothing

Submit