Evaluate 5 x − 2 y + 3 z 5 x − 2 y + 3 z when x = −2 , x = −2 , y = −4 , y = −4 , and z = 3 . z = 3 .
If you missed this problem, review Example 1.21.
Classify the equations as a conditional equation, an identity, or a contradiction and then state the solution. < − 2 x + y = −11 x + 3 y = 9 . < − 2 x + y = −11 x + 3 y = 9 .
If you missed this problem, review Example 2.6.
Classify the equations as a conditional equation, an identity, or a contradiction and then state the solution. < 7 x + 8 y = 4 3 x − 5 y = 27 . < 7 x + 8 y = 4 3 x − 5 y = 27 .
If you missed this problem, review Example 2.8.
In this section, we will extend our work of solving a system of linear equations. So far we have worked with systems of equations with two equations and two variables. Now we will work with systems of three equations with three variables. But first let's review what we already know about solving equations and systems involving up to two variables.
We learned earlier that the graph of a linear equation , a x + b y = c , a x + b y = c , is a line. Each point on the line, an ordered pair ( x , y ) , ( x , y ) , is a solution to the equation. For a system of two equations with two variables, we graph two lines. Then we can see that all the points that are solutions to each equation form a line. And, by finding what the lines have in common, we’ll find the solution to the system.
Most linear equations in one variable have one solution, but we saw that some equations, called contradictions, have no solutions and for other equations, called identities, all numbers are solutions
We know when we solve a system of two linear equations represented by a graph of two lines in the same plane, there are three possible cases, as shown.
Similarly, for a linear equation with three variables a x + b y + c z = d , a x + b y + c z = d , every solution to the equation is an ordered triple, ( x , y , z ) ( x , y , z ) , that makes the equation true.
A linear equation with three variables, where a, b, c, and d are real numbers and a, b, and c are not all 0, is of the form
a x + b y + c z = d a x + b y + c z = dEvery solution to the equation is an ordered triple, ( x , y , z ) ( x , y , z ) that makes the equation true.
All the points that are solutions to one equation form a plane in three-dimensional space. And, by finding what the planes have in common, we’ll find the solution to the system.
When we solve a system of three linear equations represented by a graph of three planes in space, there are three possible cases.
To solve a system of three linear equations, we want to find the values of the variables that are solutions to all three equations. In other words, we are looking for the ordered triple ( x , y , z ) ( x , y , z ) that makes all three equations true. These are called the solutions of the system of three linear equations with three variables .
Solutions of a system of equations are the values of the variables that make all the equations true. A solution is represented by an ordered triple ( x , y , z ) . ( x , y , z ) .
To determine if an ordered triple is a solution to a system of three equations, we substitute the values of the variables into each equation. If the ordered triple makes all three equations true, it is a solution to the system.
Determine whether the ordered triple is a solution to the system: < x − y + z = 2 2 x − y − z = −6 2 x + 2 y + z = −3 . < x − y + z = 2 2 x − y − z = −6 2 x + 2 y + z = −3 .
ⓐ ( −2 , −1 , 3 ) ( −2 , −1 , 3 ) ⓑ ( −4 , −3 , 4 ) ( −4 , −3 , 4 )