Problem 33 - Algebra 1: Concepts and Skills

Question

Use linear combinations to solve the linear system. Then check your solution. \(t+r=1\) \(2 r-t=2\)

Step-by-Step Solution

Answer

The solution to the given system of equations is \( r = 1 \) and \( t = 0 \).

Step 1: Multiply the equations

The first equation is multiplied by 2 and the second equation by 1 to set them up for combining. This gives us two new equations: \(2t + 2r = 2\) and \(2r - t = 2\).

Step 2: Add the two equations

Adding the two new equations:\( (2t + 2r) + (2r - t) = 2 + 2 \) will eliminate the term t, allowing us to solve for r.

Step 3: Simplification and Solve for r

Simplify the left hand side of equation you get:\( t + 4r = 4 \)Further divide both sides of this equation by 4, in order to isolate r:\( r = 4 / 4 = 1 \)

Step 4: Substitute \( r = 1 \) in first equation

Substitute \( r = 1 \) into the first original equation:\( t + r = 1 \)then solve for t:\( t = 1 - r = 1 - 1 = 0 \)

Step 5: Checking the solution

Place the values for \( r = 1 \) and \( t = 0 \) in the two original equations to confirm whether they are satisfied. The first equation \( t + r = 1 \) becomes \( 0 + 1 = 1 \) which is true. The second equation \( 2r - t = 2 \) becomes \( 2(1) - 0 = 2 \) which is also true. Therefore, the solution is valid.

Key Concepts

Linear Combinations

When solving linear systems, one effective method is using linear combinations. This technique involves adding or subtracting equations from each other to eliminate one variable, making it easier to solve the system.
The goal is to multiply the equations by certain numbers so that when they are added or subtracted, one variable cancels out completely.
In the exercise, the equation system is:

• \( t + r = 1 \)
• \( 2r - t = 2 \)

To demonstrate linear combinations, the first equation is multiplied by 2, resulting in:

• \( 2t + 2r = 2 \)

This setup allows for the combining of equations to strategically eliminate one variable, in this case, t.
Thus, the method of linear combinations streamlines the solving process by reducing the complexity of the system.

Solving Equations

Solving equations in a linear system involves isolating each variable step-by-step to find their specific values.
After using linear combinations, you're often left with a simpler equation ready to be solved for one of the variables.
In the given problem, adding the transformed equations yields the equation:

• \( 2t + 4r − t = 4 \)

Once simplified, this becomes:

• \( t + 4r = 4 \)

At this point, solve for one variable by isolating it. Here, solving for \( r \) gives:

• \( r = 1 \)

This is achieved after dividing both sides by 4. Then substituting \( r = 1 \) back into the original equation, \( t + r = 1 \), gives the value for t:

• \( t = 0 \)

This systematic approach ensures that each variable is correctly solved, respecting the original system's conditions.

Solution Verification

Solution verification is an indispensable step to confirm that the values obtained for variables satisfy the original system of equations.
This step ensures that no errors were made during calculations and that the solution makes sense in the context of the problem.
For the system given, we found \( r = 1 \) and \( t = 0 \). These values must now be plugged back into both original equations to check their accuracy.

• First equation: \( t + r = 1 \): \( 0 + 1 = 1 \), which holds true.
• Second equation: \( 2r - t = 2 \): \( 2(1) - 0 = 2 \), which also holds true.

If both original equations are satisfied using the found values, it reaffirms the validity of the solution.
This helps students gain confidence in their mathematical practices and ensures reliability in their answers.