Determine whether x=-7, y=5, and z=(3)/(4) is a solution to the following systems: a. ? x-3 y+4

step 1 So this system of equations is 1 minus 3. So the first equation on minus 3, 4, and then you have

Determine whether x=-7, y=5, and z=(3)/(4) is a solution to...

Key Concepts

Systems of Linear Equations

Systems of linear equations involve sets of equations with multiple variables that need to be solved simultaneously. They form the basis of many algebraic problems and are typically solved using methods such as substitution, elimination, or matrix operations, all of which aim to find variable values that satisfy every equation in the system.

Solution Verification

Solution verification is the process of checking whether a proposed set of variable values satisfies all equations in a system. This is done by substituting the values into each equation and ensuring that the equations hold true, which confirms the validity of the solution.

Substitution Method

The substitution method is a technique used to solve systems of equations by expressing one variable in terms of others from one equation and then replacing that expression in the remaining equations. This method simplifies the system, allowing each variable to be determined step by step.

Consistency of Systems

Consistency in a system of equations refers to whether there exists at least one solution that satisfies every equation simultaneously. A system is consistent if it has one or more solutions, and inconsistent if no solution exists; verifying a proposed solution helps determine the system's consistency.

Transcript

00:01 So this system of equations is 1 minus 3.

00:08 So the first equation on minus 3, 4, and then you have a minus 19 on the other side, which is the solution side, or rather the solution column.

00:20 Okay, so then for the second row, we'll have 1, 0, minus 8, equals to minus 13.

00:31 Third row we have 1, 2, 0, which is equal to 3.

00:37 So what we want to ultimately achieve is, you know, to solve this matrix is the triangle of zeros under the lower diagonal, right? so we want, this is our square matrix, right? and you have your diagonal here.

00:56 So we want zeros here.

00:58 So we want a zero here, here and here.

01:02 Ultimately okay so how are going to do this is perform row operations however we can see that we can achieve this already by swapping row 1 with row 2 okay so then this becomes 1 0 minus 8 minus 13 1 minus 3 4 minus 19 19 19 1 1 -20.

01:44 Okay, and then this will be 3.

01:48 Right.

01:48 Now, if we subject the first row from the second row, so the first, so row 2 minus row 1, also row 3 minus row 1.

02:02 So we can achieve our zeros in these positions.

02:08 This then becomes, the first row doesn't change, so it's 1 ,0 minus 0.

02:18 Minus 13 and it will be 1 minus 1 which is 0 minus 3 minus 0 which is minus 3 and 4 minus negative 8 which is 12 and negative 19 minus negative 13 will give you a negative 6 and then for row 3 minus 1 1 will be equal to 0 2 minus 0 is equal to 0.

02:50 2 minus 0 is equal to 0.

02:52 0, 8 minus negative 8, oh sorry, 0 minus negative 8 is going to be 8, and 3 minus negative 13 is going to be 16.

03:08 Okay, so now we want to achieve a 1 here.

03:13 This will make our lives easier.

03:15 So we're going to divide this row by negative 3.

03:22 So it's going to be 3 minus 1 over 3 times row 2.

03:32 And this will give us...

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