1.
Use the table of values to determine the solution of this linear system: \(4x+y=3\); \(2x+y=-5\).
The solution is where both equations have the same \(y\)-value for the same \(x\). From the table, when \(x=4\), both equations give \(y=-13\), so the solution is \((4,-13)\).
2.
In order for the lines \(6x+10y=20\) and \(y=-\frac{3}{5}x+b\) to be coincident, determine \(b\).
Convert \(6x+10y=20\) into slope-intercept form: \(10y=-6x+20\), so \(y=-\frac{3}{5}x+2\). For the two lines to be coincident, \(b=2\).
3.
The first equation is \(-6x+12y=-42\). Choose a second equation to form a system with no solution.
A system has no solution when the lines have the same slope but different \(y\)-intercepts. The first equation becomes \(y=\frac{1}{2}x-\frac{7}{2}\). Option B becomes \(y=\frac{1}{2}x+\frac{21}{2}\), so the lines are parallel and distinct.
4.
Use the graph to solve the linear system: \(y=-3x-5\), \(y-1=3x\).
The solution of a graphed linear system is the intersection point. The two lines intersect at approximately \((-1,-2)\).
5.
Model this situation with a linear system: Nate borrowed \(\$10{,}000\) for his university tuition. He borrowed part of the money at \(2.4\%\) annual interest and the rest at \(4.5\%\). His total annual interest payment is \(\$250.50\).
Let \(a\) be the amount borrowed at \(2.4\%\) and \(b\) be the amount borrowed at \(4.5\%\). The total borrowed is \(a+b=10000\). The total annual interest is \(0.024a+0.045b=250.50\).
6.
Use an elimination strategy to solve: \(4x-3y=10\); \(2x+5y=18\).
Multiply \(2x+5y=18\) by \(2\) to get \(4x+10y=36\). Subtract from \(4x-3y=10\): \(-13y=-26\), so \(y=2\). Substitute into \(4x-3y=10\): \(4x-6=10\), so \(x=4\).
7.
Which linear system is represented by the graph?
The rising line matches \(x-y=5\), since \(x-y=5\Rightarrow y=x-5\). The falling line matches \(5x+6y=18\), since \(5x+6y=18\Rightarrow y=-\frac{5}{6}x+3\).
8.
Use an elimination strategy to solve: \(3x-2y=5\); \(2x+7y=20\).
Multiply \(3x-2y=5\) by \(2\) to get \(6x-4y=10\). Multiply \(2x+7y=20\) by \(3\) to get \(6x+21y=60\). Subtract to get \(-25y=-50\), so \(y=2\). Substitute into \(3x-2y=5\): \(3x-4=5\), so \(x=3\).
9.
Determine the number of solutions of the linear system: \(5x+7y=76\); \(-25x-35y=-380\).
Put both equations into slope-intercept form. \(5x+7y=76\Rightarrow y=-\frac{5}{7}x+\frac{76}{7}\). Also, \(-25x-35y=-380\Rightarrow y=-\frac{5}{7}x+\frac{76}{7}\). Same slope and same \(y\)-intercept means the two equations represent the same line, so there are infinite solutions.
10.
Use an elimination strategy to solve: \(20x-24y=-52\); \(8x+32y=104\).
Divide the first equation by \(2\): \(10x-12y=-26\). Divide the second equation by \(4\): \(2x+8y=26\). Multiply the second by \(5\): \(10x+40y=130\). Subtract to get \(-52y=-156\), so \(y=3\). Substitute into \(20x-24y=-52\): \(20x-72=-52\), so \(x=1\).
11.
Determine the number of solutions for the linear system that models the game-points problem.
Let Player One be \(p\) and Player Two be \(q\). The two scenarios can be written as \(p=q+83\) and \(2p=2q+166\). The second equation simplifies to \(p=q+83\). The equations are the same line, so there are infinite solutions.
12.
The solution is \((-3,y)\). Determine \(y\). \(x-3y=-33\); \(\frac{6}{7}x-y=-\frac{88}{7}\).
Substitute \(x=-3\) into the first equation: \(-3-3y=-33\). Then \(-3y=-30\), so \(y=10\). The solution is \((-3,10)\).
13.
Create a linear system: Judy scored \(3\) more than twice Ann. Total \(39\) points.
Let \(j\) represent Judy's score and \(a\) represent Ann's score. Judy scored \(3\) more than twice Ann, so \(j=2a+3\). The total score is \(j+a=39\).
14.
Which linear system has solution \(x=-2\) and \(y=6\)?
Substitute \(x=-2\) and \(y=6\) into each system. In option A: \(-2+3(6)=16\) and \(4(-2)+4(6)=16\). Both equations are true.
15.
Create a linear system: nickels and dimes, four times as many dimes as nickels, total value \(\$20.25\).
Let \(d\) represent dimes and \(n\) represent nickels. Four times as many dimes as nickels gives \(d=4n\). The value equation uses cents: \(5n+10d=2025\).
16.
Car A left Calgary at 8 A.M. to travel \(500\) mi. to Regina at an average speed of \(63\) mph. Car B left Regina at the same time to travel to Calgary at \(37\) mph. The system is \(d=500-63t\) and \(d=37t\). Which graph would you use to determine how far the cars are from Regina when they meet? What is this distance?
Set the two equations equal: \(500-63t=37t\). Then \(500=100t\), so \(t=5\). Substitute into \(d=37t\): \(d=37(5)=185\). The matching graph is Graph A.
17.
Use substitution to solve: \(x=4+y\); \(4x+16y=-264\).
Substitute \(x=4+y\) into \(4x+16y=-264\): \(4(4+y)+16y=-264\). Then \(16+20y=-264\), so \(20y=-280\) and \(y=-14\). Substitute back: \(x=4+(-14)=-10\).
18.
Use substitution: perimeter of rectangular field is \(276\) m and length is \(18\) m longer than width. What are the dimensions?
Let \(l\) represent length and \(w\) represent width. Since the length is \(18\) m longer, \(l=w+18\). The perimeter equation is \(2l+2w=276\). Substitute: \(2(w+18)+2w=276\), so \(4w+36=276\), \(w=60\), and \(l=78\).
19.
Create a linear system: isosceles triangle perimeter \(36\) cm, base \(9\) cm longer than equal sides.
Let \(s\) represent each equal side and \(b\) represent the base. The perimeter is \(2s+b=36\). The base is \(9\) cm longer than each equal side, so \(s+9=b\).
20.
Match each situation to a linear system below. i) \(2l+2w=163;\ l=2w-6\). ii) \(2l+2w=163;\ w=\frac{1}{2}(l-6)\). iii) \(2l+2w=163;\ 2w=l-6\).
Situation A: length is \(6\) m less than double the width, so \(l=2w-6\), which is i. Situation B: width is one-half the length decreased by \(6\), so \(w=\frac{1}{2}(l-6)\), which is ii. Situation C: length decreased by \(6\) is double the width, so \(l-6=2w\), which is iii.
21.
Determine the number of solutions of the linear system: \(5x+7y=76\); \(-25x-35y=-380\).
Put both equations into slope-intercept form. \(5x+7y=76\Rightarrow y=-\frac{5}{7}x+\frac{76}{7}\). Also, \(-25x-35y=-380\Rightarrow y=-\frac{5}{7}x+\frac{76}{7}\). Same slope and same \(y\)-intercept means the two equations represent the same line, so there are infinite solutions.
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