1.5.1: Exercises 1.5
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In Exercises \(\PageIndex{1}\) - \(\PageIndex{5}\), find the solution of the given problem by:
- creating an appropriate system of linear equations
- forming the augmented matrix that corresponds to this system
- putting the augmented matrix into reduced row echelon form
- interpreting the reduced row echelon form of the matrix as a solution
A farmer looks out his window at his chickens and pigs. He tells his daughter that he sees 62 heads and 190 legs. How many chickens and pigs does the farmer have?
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29 chickens and 33 pigs
A lady buys 20 trinkets at a yard sale. The cost of each trinket is either $0.30 or $0.65. If she spends $8.80, how many of each type of trinket does she buy?
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12 $0.30 trinkets, 8 $0.65 trinkets
A carpenter can make two sizes of table, grande and venti. The grande table requires 4 table legs and 1 table top; the venti requires 6 table legs and 2 table tops. After doing work, he counts up spare parts in his warehouse and realizes that he has 86 table tops left over, and 300 legs. How many tables of each kind can he build and use up exactly all of his materials?
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42 grande tables, 22 venti tables
A jar contains 100 marbles. We know there are twice as many green marbles as red; that the number of blue and yellow marbles together is the same as the number of green; and that three times the number of yellow marbles together with the red marbles gives the same numbers as the blue marbles. How many of each color of marble are in the jar?
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35 blue, 40 green, 20 red, 5 yellow
A rescue mission has 85 sandwiches, 65 bags of chips and 210 cookies. They know from experience that men will eat 2 sandwiches, 1 bag of chips and 4 cookies; women will eat 1 sandwich, a bag of chips and 2 cookies; kids will eat half a sandwhich, a bag of chips and 3 cookies. If they want to use all their food up, how many men, women and kids can they feed?
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30 men, 15 women, 20 kids
In Exercises \(\PageIndex{6}\) - \(\PageIndex{15}\), find the polynomial with the smallest degree that goes through the given points.
\((1,3)\) and \((3,15)\)
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\(f(x) = 6x-3\)
\((-2,14)\) and \((3,4)\)
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\(f(x) = -2x+10\)
\((1,5)\), \((-1,3)\) and \((3,-1)\)
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\(f(x) = -x^2+x+5\)
\((-4,-3)\), \((0,1)\) and \((1,4.5)\)
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\(f(x) = \frac12x^2+3x+1\)
\((-1,-8)\), \((1,-2)\) and \((3,4)\)
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\(f(x) = 3x-5\)
\((-3,3)\), \((1,3)\) and \((2,3)\)
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\(f(x) = 3\)
\((-2,15)\), \((-1,4)\), \((1,0)\) and \((2,-5)\)
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\(f(x) = -x^3+x^2-x+1\)
\((-2,-7)\), \((1,2)\), \((2,9)\) and \((3,28)\)
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\(f(x) = x^3+1\)
\((-3,10)\), \((-1,2)\), \((1,2)\) and \((2,5)\)
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\(f(x) = x^2+1\)
\((0,1)\), \((-3,-3.5)\), \((-2,-2)\) and \((4,7)\)
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\(f(x) = \frac32x+1\)
The general exponential function has the form \(f(x)= ae^{bx}\), where \(a\) and \(b\) are constants and \(e\) is Euler’s constant (\(\approx\) 2.718). We want to find the equation of the exponential function that goes through the points \((1,2)\) and \((2,4)\).
- Show why we cannot simply substitute in values for \(x\) and \(y\) in \(y = ae^{bx}\) and solve using the techniques we used for polynomials.
- Show how the equality \(y = ae^{bx}\) leads us to the linear equation \(\ln y = \ln a + bx\).
- Use the techniques we developed to solve for the unknowns \(\ln a\) and \(b\).
- Knowing \(\ln a\), find \(a\); find the exponential function \(f(x) = ae^{bx}\) that goes through the points \((1,2)\) and \((2,4)\).
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- Substitution yields the equations \(2 = ae^{b}\) and \(4 = ae^{2b}\); these are not linear equations.
- \(y=ae^{bx}\) implies that \(\ln y = \ln (ae^{bx}) = \ln a + \ln e^{bx} = \ln a + bx\).
- Plugging in the points for \(x\) and \(y\) in the equation \(\ln y = \ln a + bx\), we have equations \[\begin{array}{ccccc} \ln a & + & b & = & \ln 2 \\ \ln a & + & 2b & = & \ln 4\\ \end{array}. \nonumber \] To solve,
\[\left[\begin{array}{ccc}{1}&{1}&{\ln 2}\\{1}&{2}&{\ln 4}\end{array}\right]\qquad\overrightarrow{\text{rref}}\qquad\left[\begin{array}{ccc}{1}&{0}&{0}\\{0}&{1}&{\ln 2}\end{array}\right] \nonumber \] Therefore \(\ln a = 0\) and \(b = \ln 2\). - Since \(\ln a = 0\), we know that \(a = e^0 = 1\). Thus our exponential function is \(f(x) = e^{x\ln 2}\).
In a football game, 24 points are scored from 8 scoring occasions. The number of successful extra point kicks is equal to the number of successful two point conversions. Find all ways in which the points may have been scored in this game.
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The augmented matrix from this system is \(\left[\begin{array}{ccccc}{1}&{1}&{1}&{1}&{8}\\{6}&{1}&{2}&{3}&{24}\\{0}&{1}&{-1}&{0}&{0}\end{array}\right]\) From this we find the solution \[\begin{align}\begin{aligned} t&=\frac83-\frac13f\\ x&=\frac83-\frac13f\\ w&=\frac83-\frac13f.\end{aligned}\end{align} \nonumber \] The only time each of these variables are nonnegative integers is when \(f=2\) or \(f=8\). If \(f=2\), then we have 2 touchdowns, 2 extra points and 2 two point conversions (and 2 field goals); this doesn’t make sense since the extra points and two point conversions follow touchdowns. If \(f=8\), then we have no touchdowns, extra points or two point conversions (just 8 field goals). This is the only solution; all points were scored from field goals.
In a football game, 29 points are scored from 8 scoring occasions. There are 2 more successful extra point kicks than successful two point conversions. Find all ways in which the points may have been scored in this game.
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The augmented matrix from this system is \(\left[\begin{array}{ccccc}{1}&{1}&{1}&{1}&{8}\\{6}&{1}&{2}&{3}&{29}\\{0}&{1}&{-1}&{0}&{2}\end{array}\right]\). From this we find the solution \[\begin{align}\begin{aligned} t&=4-\frac13f\\ x&=3-\frac13f\\ w&=1-\frac13f.\end{aligned}\end{align} \nonumber \] The only time each of these variables are nonnegative integers is when \(f=0\) or \(f=3\). If \(f=0\), then we have 4 touchdowns, 3 extra points and 1 two point conversions (no field goals). If \(f=3\), then we have 3 touchdowns, 2 extra points and no two point conversions (and 3 field goals).
In a basketball game, where points are scored either by a 3 point shot, a 2 point shot or a 1 point free throw, 80 points were scored from 30 successful shots. Find all ways in which the points may have been scored in this game.
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Let \(x_1\), \(x_2\) and \(x_3\) represent the number of free throws, 2 point and 3 point shots taken. The augmented matrix from this system is \(\left[\begin{array}{cccc}{1}&{1}&{1}&{30}\\{1}&{2}&{3}&{80}\end{array}\right]\). From this we find the solution \[\begin{align}\begin{aligned} x_1&=-20+x_3\\ x_2&=50-2x_3.\end{aligned}\end{align} \nonumber \] In order for \(x_1\) and \(x_2\) to be nonnegative, we need \(20\leq x_3\leq 25\). Thus there are 6 different scenerios: the “first” is where 20 three point shots are taken, no free throws, and 10 two point shots; the “last” is where 25 three point shots are taken, 5 free throws, and no two point shots.
In a basketball game, where points are scored either by a 3 point shot, a 2 point shot or a 1 point free throw, 110 points were scored from 70 successful shots. Find all ways in which the points may have been scored in this game.
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Let \(x_1\), \(x_2\) and \(x_3\) represent the number of free throws, 2 point and 3 point shots taken. The augmented matrix from this system is \(\left[\begin{array}{cccc}{1}&{1}&{1}&{70}\\{1}&{2}&{3}&{110}\end{array}\right]\). From this we find the solution \[\begin{align}\begin{aligned} x_1&=30+x_3\\ x_2&=40-2x_3.\end{aligned}\end{align} \nonumber \] In order for \(x_2\) to be nonnegative, we need \(x_3\leq 20\). Thus there are 21 different scenerios: the “first” is where 0 three point shots are taken (\(x_3=0\), 30 free throws and 40 two point shots; the “last” is where 20 three point shots are taken, 50 free throws, and no two point shots.
Describe the equations of the linear functions that go through the point (1,3). Give 2 examples.
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Let \(y = ax+b\); all linear functions through (1,3) come in the form \(y = (3-b)x+b\). Examples: \(b=0\) yields \(y = 3x\); \(b=2\) yields \(y=x+2\).
Describe the equations of the linear functions that go through the point (2,5). Give 2 examples.
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Let \(y = ax+b\); all linear functions through (2,5) come in the form \(y = (2.5-\frac12b)x+b\). Examples: \(b=1\) yields \(y = 2x+1\); \(b=-1\) yields \(y=3x-1\).
Describe the equations of the quadratic functions that go through the points \((2,-1)\) and (1,0). Give 2 examples.
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Let \(y = ax^2+bx+c\); we find that \(a = -\frac12+\frac12 c\) and \(b = \frac12-\frac32c\). Examples: \(c=1\) yields \(y = -x+1\); \(c=3\) yields \(y=x^2-4x+3\).
Describe the equations of the quadratic functions that go through the points \((-1,3)\) and (2,6). Give 2 examples.
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Let \(y = ax^2+bx+c\); we find that \(a = 2-\frac12 c\) and \(b = -1+\frac12c\). Examples: \(c=0\) yields \(y = 2x^2-x\); \(c=-2\) yields \(y=3x^2-2x-2\).