# system of equations problems 3 variables

Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. A system in upper triangular form looks like the following: \begin{align*} Ax+By+Cz &= D \nonumber \\[4pt] Ey+Fz &= G \nonumber \\[4pt] Hz &= K \nonumber \end{align*} \nonumber. He earned $$670$$ in interest the first year. Just as with systems of equations in two variables, we may come across an inconsistent system of equations in three variables, which means that it does not have a solution that satisfies all three equations. \begin{align}y+2z&=3 \\ -y-z&=-1 \\ \hline z&=2 \end{align}\hspace{5mm}\begin{align}(4)\5)\\(6)\end{align}. A system of three equations in three variables can be solved by using a series of steps that forces a variable to be eliminated. Systems of equations in three variables that are inconsistent could result from three parallel planes, two parallel planes and one intersecting plane, or three planes that intersect the other two but not at the same location. (b) Three planes intersect in a line, representing a three-by-three system with infinite solutions. Solve this system using the Addition/Subtraction method. Example \(\PageIndex{5}: Finding the Solution to a Dependent System of Equations. Call the changed equations … Example: At a store, Mary pays 34 for 2 pounds of apples, 1 pound of berries and 4 pounds of cherries. Add equation (2) to equation (3) and write the result as equation (3). You can visualize such an intersection by imagining any corner in a rectangular room. This will be the sample equation used through out the instructions: Equation 1) x – 6y – 2z = -8. Use the resulting pair of equations from steps 1 and 2 to eliminate one of the two remaining variables. The final equation $0=2$ is a contradiction, so we conclude that the system of equations in inconsistent and, therefore, has no solution. Solve the system of three equations in three variables. \begin{align}−5x+15y−5z&=−20 \\ 5x−13y+13z&=8 \\ \hline 2y+8z&=−12\end{align}\hspace{5mm} \begin{align}&(1)\text{ multiplied by }−5 \\ &(3) \\ &(5) \end{align}. Adding equations (1) and (3), we have, \begin{align}2x+y−3z=0 \\ x−y+z=0 \\ \hline 3x−2z=0 \end{align}. \begin{align}y+2\left(2\right)&=3 \\ y+4&=3 \\ y&=-1 \end{align}. $\begin{array}{rrr} { \text{} \nonumber \\[4pt] x+y+z=2 \nonumber \\[4pt] (3)+(−2)+(1)=2 \nonumber \\[4pt] \text{True}} & {6x−4y+5z=31 \nonumber \\[4pt] 6(3)−4(−2)+5(1)=31 \nonumber \\[4pt] 18+8+5=31 \nonumber \\[4pt] \text{True} } & { 5x+2y+2z = 13 \nonumber \\[4pt] 5(3)+2(−2)+2(1)=13 \nonumber \\[4pt] 15−4+2=13 \nonumber \\[4pt] \text{True}} \end{array}$. John invested2,000 in a money-market fund, $3,000 in municipal bonds, and$7,000 in mutual funds. $\begin{array}{l}\text{ }x+y+z=2\hfill \\ \text{ }y - 3z=1\hfill \\ 2x+y+5z=0\hfill \end{array}$. To make the calculations simpler, we can multiply the third equation by 100. Solve the system created by equations (4) and (5). This will change equations (1) and (2) to equations in the two variables and . In "real life", these problems can be incredibly complex. In this system, each plane intersects the other two, but not at the same location. Access these online resources for additional instruction and practice with systems of equations in three variables. But let’s say we have the following situation. Solve the resulting two-by-two system. Or two of the equations could be the same and intersect the third on a line. Back-substitute that value in equation (2) and solve for $$y$$. At the end of the year, she had made 1,300 in interest. Problem 3.1b: The standard equation of a circle is x 2 +y 2 +Ax+By+C=0. In this solution, $x$ can be any real number. Determine whether the ordered triple $\left(3,-2,1\right)$ is a solution to the system. Finally, we can back-substitute $$z=2$$ and $$y=−1$$ into equation (1). Graphically, a system with no solution is represented by three planes with no point in common. The solution is x = –1, y = 2, z = 3. 2) Now, solve the two resulting equations (4) and (5) and find the value of x and y . We can solve for $$z$$ by adding the two equations. \begin{align} x+y+z=2\\ \left(3\right)+\left(-2\right)+\left(1\right)=2\\ \text{True}\end{align}\hspace{5mm} \hspace{5mm}\begin{align} 6x - 4y+5z=31\\ 6\left(3\right)-4\left(-2\right)+5\left(1\right)=31\\ 18+8+5=31\\ \text{True}\end{align}\hspace{5mm} \hspace{5mm}\begin{align}5x+2y+2z=13\\ 5\left(3\right)+2\left(-2\right)+2\left(1\right)=13\\ 15 - 4+2=13\\ \text{True}\end{align}. Looking at the coefficients of $$x$$, we can see that we can eliminate $$x$$ by adding Equation \ref{4.1} to Equation \ref{4.2}. Solving 3 variable systems of equations with no or infinite solutions. Solve the resulting two-by-two system. The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. Equation 3) 3x - 2y – 4z = 18 STEP Use the linear combination method to rewrite the linear system in three variables as a linear system in twovariables. John invested4,000 more in municipal funds than in municipal bonds. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Write the result as row 2. Choose another pair of equations and use them to eliminate the same variable. The total interest earned in one year was $$670$$. See Example $$\PageIndex{1}$$. Pick another pair of equations and solve for the same variable. Infinite number of solutions of the form $$(x,4x−11,−5x+18)$$. Looking at the coefficients of $x$, we can see that we can eliminate $x$ by adding equation (1) to equation (2). \begin{align}x+y+z=12{,}000\hfill \\ 3x+4y +7z=67{,}000 \\ -y+z=4{,}000 \end{align}. In the problem posed at the beginning of the section, John invested his inheritance of 12,000 in three different funds: part in a money-market fund paying 3% interest annually; part in municipal bonds paying 4% annually; and the rest in mutual funds paying 7% annually. How to solve a word problem using a system of 3 equations with 3 variable? 2. Thus, \begin{align} x+y+z &=12,000 \; &(1) \nonumber \\[4pt] −y+z &= 4,000 \; &(2) \nonumber \\[4pt] 3x+4y+7z &= 67,000 \; &(3) \nonumber \end{align} \nonumber. Find the solution to the given system of three equations in three variables. The third equation shows that the total amount of interest earned from each fund equals $$670$$. 3) Substitute the value of x and y in any one of the three given equations and find the value of z . Solve for $$z$$ in equation (3). How much did he invest in each type of fund? Next, we back-substitute $$z=2$$ into equation (4) and solve for $$y$$. After performing elimination operations, the result is an identity. John invested $$4,000$$ more in municipal funds than in municipal bonds. Example 2. So, let’s first do the multiplication. Step 2: Substitute this value for in equations (1) and (2). You have created a system of two equations in two unknowns. Systems that have a single solution are those which, after elimination, result in a. Understanding the correct approach to setting up problems such as this one makes finding a solution a matter of following a pattern. Then, we multiply equation (4) by 2 and add it to equation (5). How much did he invest in each type of fund? Problem : Solve the following system using the Addition/Subtraction method: 2x + y + 3z = 10. The calculator will use the Gaussian elimination or Cramer's rule to generate a step by step explanation. This calculator solves system of three equations with three unknowns (3x3 system). First, we assign a variable to each of the three investment amounts: \begin{align} x &= \text{amount invested in money-market fund} \nonumber \\[4pt] y &= \text{amount invested in municipal bonds} \nonumber \\[4pt] z &= \text{amount invested in mutual funds} \nonumber \end{align} \nonumber. The solutions to systems of equations are the variable mappings such that all component equations are satisfied—in other words, the locations at which all of these equations intersect. You’re going to the mall with your friends and you have200 to spend from your recent birthday money. Systems of three equations in three variables are useful for solving many different types of real-world problems. The final equation $$0=2$$ is a contradiction, so we conclude that the system of equations in inconsistent and, therefore, has no solution. Improve your math knowledge with free questions in "Solve a system of equations in three variables using elimination" and thousands of other math skills. The same is true for dependent systems of equations in three variables. A system of equations is a set of one or more equations involving a number of variables. 1. Then, we write the three equations as a system. Unless it is given, translate the problem into a system of 3 equations using 3 variables. The first equation indicates that the sum of the three principal amounts is $$12,000$$. However, finding solutions to systems of three equations requires a bit more organization and a touch of visual gymnastics. The third equation can be solved for $$z$$,and then we back-substitute to find $$y$$ and $$x$$. Lee Pays \$49 for 5 pounds of apples, 3 pounds of berries, and 2 pounds of cherries. In your studies, however, you will generally be faced with much simpler problems.

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