Systems of Linear Equations

Systems of Equations

A system of equations consists of two or more equations, considered as simultaneous specifications on more than one variable. A solution to a system of equations is an ordered assignment of values of the variables that, when substituted, would make each of the equations into true statements. The process of finding the solutions of a system is called solving the system. The set of all solutions is called the solution set of the system. Systems with the same solution set are called equivalent systems.

EXAMPLE 30.1 Verify that $ (x,y)=(-4,2) $ is a solution to the system

\[ y^{2}+x=0 \]

\[ 2x+3y=-2 \]

If x = -4 and y = 2, then equation (1) becomes $ 2^{2} + (-4) = 0 $ and equation (2) becomes $ 2(-4) + 3 = -2 $ . Since these are both true statements, $ (x, y) = (-4, 2) $ is a solution to the system.

Systems of Linear Equations

A linear equation in several variables $ x_{1}x_{2},,x_{n} $ is one that can be written in the form $ a_{1}x_{1}+a_{2}x_{2}++a_{n}x_{n}=b $ , where the $ a_{i} $ are constants. This is referred to as standard form. If all equations of a system are linear, the system is called a linear system; if all equations are in standard form, the system is also considered to be in standard form.

EXAMPLE 30.2 Rewrite the system

\[ \begin{aligned}2x+4y&=5x-6y\\y+5&=3x+5y\end{aligned} \]

in standard form.

An equation in standard form must have all variable terms on the left side and any constant terms on the right side. Here equation (1) violates the first of these conditions and equation (2) violates both. Hence, add $ -5x + 6y $ to both sides of equation (1) to obtain $ -3x + 10y = 0 $ , and add -3x - 5y and -5 to both sides of equation (2) to obtain -3x - 4y = -5. The resulting equations are in standard form:

\[ \begin{aligned}-3x+10y&=0\\-3x-4y&=-5\end{aligned} \]

Equivalent Systems

Equivalent systems of linear equations can be produced by the following operations on equations. (It is understood that “adding two equations” means adding left side to left side and right side to right side to produce a new equation and “multiple of an equation” means the result of multiplying left side and right side by the same constant.)

  1. Interchanging two equations.

  2. Replacing an equation by a nonzero multiple of itself.

  3. Replacing an equation by the result of adding the equation to a multiple of another equation.

EXAMPLE 30.3 For the system of the previous example, find an equivalent system in which one equation does not contain the variable x.

If both sides of equation (4) are multiplied by -1, the coefficient of x will be the opposite of the coefficient of x in equation (3). Hence, replacing equation (3) by itself added to -1 times equation (4) will achieve the required result:

\[ 14y=5 \]

\[ -3x-4y=-5 \]

Classification of Linear Systems

It is shown in advanced courses that systems of linear equations fall into one of three categories:

  1. CONSISTENT AND INDEPENDENT. Such systems have exactly one solution.

  2. INCONSISTENT. Such systems have no solutions.

  3. DEPENDENT. Such systems have an infinite number of solutions.

Solutions of Linear Systems in Two Variables

Solutions of linear systems in two variables are found by three methods:

  1. GRAPHICAL METHOD. Graph each equation (each graph is a straight line). If the lines intersect in a single point, the coordinates of this point may be read from the graph. After checking by substitution in each equation, these coordinates are the solution of the system. If the lines coincide, the system is dependent, and there are an infinite number of solutions, with each solution to one equation being a solution of the others. If neither of these situations occurs, the system is inconsistent.

  2. SUBSTITUTION METHOD. Solve one equation for one variable in terms of the other. Substitute this expression into the other equations to determine the value of the first variable (if possible). Then substitute this value to determine the value of the other variable.

  3. ELIMINATION METHOD. Apply the operations on equations leading to equivalent systems to eliminate one variable from one equation, solve the resulting equation for this variable, and substitute this value to determine the value of the other variable.

In methods 2 and 3, the occurrence of an equation of the form a = b, where a and b are unequal constants, indicates an inconsistent system. If this does not occur, but all equations except one reduce to 0 = 0, the system is dependent, and there are an infinite number of solutions, with each solution of one equation being a solution of the others.

Solutions of Linear Systems in More than Two Variables

Solutions of linear systems in more than two variables are found by two methods:

  1. SUBSTITUTION METHOD. Solve one equation for one variable in terms of the others. Substitute this expression into the other equations to obtain a system with one fewer variable. If this process can be continued until an equation in one variable is obtained, solve the resulting equation for this variable, and substitute this value to determine the value of the other variables.

  2. ELIMINATION METHOD. Apply the operations on equations leading to equivalent systems to eliminate one variable from all equations except one. This leads to a system with one fewer variable. If this process can be continued until an equation in one variable is obtained, solve the resulting equation for this variable, and substitute this value to determine the value of the other variables.

Again, the occurrence of an equation of the form a = b, where a and b are unequal constants, indicates an inconsistent system. If this does not occur, but one or more equations reduce to 0 = 0, leaving fewer nontrivial equations than there are variables, the system is dependent, and there are an infinite number of solutions, with each solution of one equation being a solution of the others.

SOLVED PROBLEMS

30.1. Solve the system $ \[\begin{array}{r l r}{2x+3y=6}&{{}}&{(1)}\\ {-3x-y=5}&{{}}&{(2)}\end{array}\]

$ (a) graphically; (b) by substitution; (c) by elimination.

  1. Graph the two equations in the same Cartesian coordinate system (Fig. 30-1); the graphs are straight lines.

Figure 30-1

The two lines appear to intersect at $ (-3,4) $ . It is necessary to check this result: substituting x = -3 and y = 4 into equations (1) and (2) yields

\[ \begin{aligned}&2(-3)+3\cdot4=6\quad&-3(-3)-4=5\\&&and\quad&\\&6=6\quad&5=5\end{aligned} \]

respectively. Thus $ (-3,4) $ is the only solution of the system.

  1. It is correct to begin by solving either equation for either variable in terms of the other. The simplest choice seems to be to solve equation (2) for y in terms of x to obtain

\[ y=-3x-5 \]

Substitute the expression -3x - 5 for y into equation (1) to obtain

\[ \begin{aligned}2x+3(-3x-5)&=6\\-7x-15&=6\\-7x&=21\\x&=-3\end{aligned} \]

Substitute -3 for x into equation (2) to obtain

\[ \begin{aligned}-3(-3)-y&=5\\9-y&=5\\y&=4\end{aligned} \]

Again, $ (-3,4) $ is the only solution of the system.

  1. If equation (2) is multiplied by 3, the coefficient of y will “match” the coefficient of y in equation (1); that is, it will be equal in absolute value and opposite in sign. Equation (2) then becomes

\[ -9x-3y=15 \]

If equation (1) is replaced by itself plus this multiple of equation (2), the following equivalent system results:

\[ \begin{aligned}-7x&=21\\-3x-y&=5\end{aligned} \]

From equation (4), x = -3. Substituting into equation (2) yields y = 4, as before.

30.2. Solve the system $ \[\begin{array}{r l r l} y &= 2x + 2 & (1) \\ 4x - 2y &= 8 & (2) \end{array}\]

$ (a) graphically; (b) nongraphically.

  1. Graph the two equations in the same Cartesian coordinate system (Fig. 30-2); the graphs are straight lines.

Figure 30-2

The lines appear to be parallel. In fact, since both have slope 2, but different y intercepts, the lines are parallel; there is no point of intersection, and the system has no solution (inconsistent system).

  1. Solve by substitution: Substitute the expression $ 2x + 2 $ for y from equation (1) into equation (2).

\[ \begin{aligned}4x-2(2x+2)&=8\\4x-4x-4&=8\\-4&=8\end{aligned} \]

Thus there is no solution, and the system is inconsistent.

30.3. Solve the system $ \[\begin{array}{r}4x + 2y = 6 \\ 6x + 3y = 9\end{array}\]

$ (1) graphically; (b) nongraphically. (2) (a) graphically; (b) nongraphically.

  1. Graph the two equations in the same Cartesian coordinate system (Fig. 30-3); the graphs are straight lines.

Figure 30-3

The lines appear to coincide. In fact, since both have slope -2 and y-intercept 3, they do coincide. The system is dependent; every solution of one equation is a solution of the other equation. All solutions can be summarized as follows:

  1. Let y = c, where c is any real number. Then substituting c for y into one equation, say (1), and solving for x yields:

\[ \begin{aligned}4x+2c&=6\\4x&=6-2c\\x&=\frac{3-c}{2}\end{aligned} \]

Hence all solutions of the system can be written as $ (, c) $ , where c is any real number.

30.4. Show in tabular form the algebraic and geometric interpretations of the types of systems of linear equations in two variables.

The systems are characterized as consistent and independent, inconsistent, or dependent. Algebraic interpretation means the number of solutions; geometric interpretation means the behavior of the graphs.

Type of System Number of Solutions Behavior of Graphs
Consistent and independent One Lines intersect in one point
Inconsistent None Two lines are parallel, more than two lines fail to intersect in one point
Dependent Infinite Lines coincide

30.5. Solve the system

\[ \begin{aligned}x-3y+2z&=14\\2x+5y-z&=-9\\-3x-y+2z&=2\end{aligned} \]

    1. by substitution; (b) by elimination.
  1. Solve equation (1) for x to obtain

\[ x=3y-2z+14 \]

Substitute the expression $ 3y - 2z + 14 $ for x from equation (4) into equations (2) and (3).

\[ \begin{aligned}2(3y-2z+14)+5y-z&=-9\\-3(3y-2z+14)-y+2z&=2\end{aligned} \]

Simplifying yields:

\[ \begin{aligned}11y-5z&=-37\\-10y+8z&=44\end{aligned} \]

Solve equation (5) for y to obtain

\[ y=\frac{5z-37}{11} \]

Substitute the expression on the right for y into equation (6).

\[ \begin{aligned}-10\Big(\frac{5z-37}{11}\Big)+8z&=44\\-50z+370+88z&=484\\38z&=114\\z&=3\end{aligned} \]

Substituting this value for z into equation (7) yields y = -2. Substituting y = -2 and z = 3 into equation (4) yields x = 2. There is exactly one solution, written as an ordered triple $ (2, -2, 3) $ .

  1. Replacing equation (2) by itself plus -2 times equation (1) will eliminate x from equation (2).

Thus:

\[ \begin{aligned}&2x+5y-z=-9\quad&(2)\\\frac{-2x+6y-4z=-28}{11y-5z=-37}\quad&\begin{array}{c}(-2)\cdot Eq.(1)\ $ 5)\end{array}\end{aligned} \]

Similarly, replacing equation (3) by itself plus 3 times equation (1) will eliminate x from equation (3):

\[ \begin{aligned}&-3x-y+2z=2\quad&(3)\\&\frac{3x-9y+6z=42}{-10y+8z=44}\quad&3\cdot Eq.(1)\\\end{aligned} \]

Solving the system (5), (6) by elimination yields the same solution as above: (2, -2, 3).

\[ x-4y-5z=8\qquad(1) \]

30.6. Solve the system 4x - 2z = 10 (2).

Replace equation (2) by itself -4 times equation (1):

\[ \begin{aligned}&\frac{\begin{aligned}\\ &4x&-2z&=10\\&-4x+16y+20z&=-32\\&16y+18z&=-22\\ &\end{aligned}}{\begin{aligned}\\ &(-4)\cdot Eq.(1)\\&(4)\\ &\end{aligned}}\\ \end{aligned} \]

Replace equation (3) by itself plus -5 times equation (1).

\[ \begin{aligned}&\frac{5x\quad-4y\quad-7z=3}{\begin{array}{c}5x+20y+25z=-40\\16y+18z=-37\end{array}}\quad\begin{array}{c}(3)\ $ -5)\cdot Eq.(1)\ $ 5)\end{array}\end{aligned} \]

The system (1), (4), (5) is clearly inconsistent, since adding -1 times equation (4) to equation (5) yields 0 = -15. Thus there is no solution.

30.7. Solve the system 2x - 2y - 10z = -6

\[ \begin{aligned}x+y+z&=1\\2x-2y-10z&=-6\\-x+3y+11z&=7\end{aligned} \]

Replace equation (2) by itself plus -2 times equation (1):

\[ \begin{aligned}&2x-2y-10z=-6\quad&(2)\\-2x-2y-2z=-2\quad&(-2)\cdot Eq.(1)\\-4y-12z=-8\quad&(4)\end{aligned} \]

Replace equation (3) by itself plus equation (1):

\[ \begin{aligned}\frac{-x+3y+11z=7\quad&(3)\\\frac{x+y+z=1\quad&(1)\\4y+12z=8\quad&(5)\end{aligned} \]

The system (1), (4), (5)

\[ \begin{aligned}x+y+z&=1\quad&(1)\\-4y-12z&=-8\quad&(4)\\4y+12z&=8\quad&(5)\end{aligned} \]

is clearly dependent, since replacing equation (5) by itself plus equation (4) yields

\[ \begin{aligned}x+y+z&=1\quad&(1)\\-4y-12z&=-8\quad&(4)\\0&=0\quad&(6)\end{aligned} \]

Thus there is an infinite number of solutions. To express them all, let z = c, c any real number. Then solving -4y - 12c = -8 for y yields y = 2 - 3c. Substituting y = 2 - 3c and z = c into equation (1) yields

\[ x+2-3c+c=1 \]

\[ x=2c-1 \]

Thus all solutions can be written as ordered triples \((2c-1,2-3c,c)\), \(c\) any real number.

30.8. $8000 is to be invested, part at 6% interest, and part at 11% interest. How much should be invested at each rate if a total return of 9% is desired?

Use the formula \(I = Prt\) with \(t\) understood to be one year. Let \(x = \text{amount invested}\) at \(6\%\) and \(y = \text{amount invested}\) at \(11\%\); a tabular arrangement is helpful:

P: Amount Invested r: Rate of Interest I: Interest Earned
First account x 0.06 0.06x
Second account y 0.11 0.11y
Total investment 8000 0.09 0.09(8000)

Since the amounts invested add up to the total investment,

\[ x+y=8000\qquad(1) \]

Since the interest earned adds up to the total interest,

\[ 0.06x+0.11y=0.09(8000) \]

The system (1), (2) can be solved by elimination. Replace equation (2) by itself plus -0.06 times equation (1):

\[ \begin{aligned}\frac{\begin{array}{c}0.06x+0.11y=0.09(8000)\\-0.06x-0.06y=-0.06(8000)\\0.05y=0.03(8000\\\end{array}}{(2)}\quad&\begin{array}{c}(-0.06)\cdot Eq.(1)\ $ 3)\\\end{array}\end{aligned} \]

Hence y = 4800. Substituting into equation (1) yields x = 3200, hence $3200 should be invested at 6% and $4800 at 11%.

30.9. Find a, b, and c so that the graph of the circle with equation $ x^{2} + y^{2} + ax + by + c = 0 $ passes through the points (1,5), (4,4), and (3,1).

If a point lies on the graph of an equation, the coordinates of the point satisfy the equation. Hence, substitute, in turn, \((x, y) = (1, 5)\), \((x, y) = (4, 4)\), and \((x, y) = (3, 1)\) to obtain

\[ 1+25+a1+b5+c=0 \]

\[ a+5b+c=-26 \]

\[ 16+16+a4+b4+c=0\qquad or,simplifying\qquad4a+4b+c=-32 \]

\[ 9+1+a3+b1+c=0 \]

\[ 3a+b+c=-10 \]

To solve the system (1), (2), (3), eliminate a from equations (2) and (3) as follows:

\[ a+5b+c=-26 \]

\[ -16b-3c=72 \]

\[ (4)=\mathrm{Eq.}\left(2\right)+\left(-4\right)\cdot\mathrm{Eq.}\left(1\right) \]

\[ -14b-2c=68 \]

\[ (5)=\mathrm{Eq.}\left(3\right)+\left(-3\right)\cdot\mathrm{Eq.}\left(1\right) \]

Now eliminate b from equation (5) by replacing it with itself plus -7/8 times equation (4).

\[ a+5b+c=-26 \]

\[ -16b-3c=72 \]

\[ \frac{5}{8}c=5 \]

Finally, solve equation (6) to obtain c = 8 and substitute in turn into equations (4) and (1) to obtain b = -6 and a = -4.

The equation of the circle is $ x^{2} + y^{2} - 4x - 6y + 8 = 0 $ .

SUPPLEMENTARY PROBLEMS

30.10. Solve the systems (a) $ \[\begin{array}{r}2x - 3y = 4 \\ 3x + 2y = 19\end{array}\]

$

  1. $

    \[\begin{array}{r}6x - 4y = 8 \\ 9x - 6y = 12\end{array}\]

    $

  2. $

    \[\begin{array}{r}2y = 3x + 4 \\ 9x - 6y = 4\end{array}\]

    $

Ans. (a) (5, 2); (b) $ (, c) $ , c any real number; (c) no solution

30.11. Solve the systems (a)

\[ \begin{aligned}3x-2y&=0\\x+3y&=0\\2x-y&=0\end{aligned} \]

\[ \begin{aligned}x-3y&=0\\ 2x+3y&=2\\ -x+y&=1\end{aligned} \]

\[ x+2y=2 \]

  1. 2x - y = 3

\[ 3x+y=5 \]

Ans. (a) (0, 0); (b) no solution; (c) $ (, ) $

30.12. Solve the systems:

\[ \begin{aligned}x+y+z&=5\\ (a)\quad x-4y-3z&=11\\ -2x+2y+5z&=-30\end{aligned} \]

\[ \begin{aligned}x+y-2z&=4\\2x-5y+z&=7\\x+8y-7z&=2\end{aligned} \]

\[ \begin{aligned}-x+2y+2z&=-13\ $ c)\quad5x+y-8z&=0\\3x-y&=12\end{aligned} \]

Ans. (a) $ (7,2,-4) $ ; (b) no solution; (c) $ (2,-6,) $

2x - y - z = 0

30.13. Solve the systems (a) $ x - y + z = 0 $

$ 3x + 2y + z = 0 $

\[ \begin{aligned}x+y-z&=5\\3x-y+z&=3\\y-z&=3\end{aligned} \]

\[ \begin{aligned}(c)\quad&3x-3y-6z=-15\\&-2x+2y+4z=10\end{aligned} \]

Ans. (a) $ (0, 0, 0) $ ; (b) $ (2, 3 + c, c) $ , c any real number; (c) $ (c + 2d - 5, c, d) $ , c and d any real numbers

30.14. $16,500 was invested in three accounts, yielding an annual return of 5%, 8%, and 10%, respectively. The amount invested at 5% was equal to the amount invested at 8% plus twice the amount invested at 10%. How much was invested at each rate if the total return on the investment was $1085?

Ans. $9500 at 5%, $4500 at 8%, $2500 at 10%

30.15. Find \(a, b\), and \(c\) so that the equation of the parabola \(y = ax^{2} + bx + c\) passes through \((1,4)\), \((-1,6)\), and \((2,12)\).

Ans. \(a = 3\), \(b = -1\), \(c = 2\)