1.
The first equation of a linear system is \(8x+13y=166\). Choose a second equation to form a linear system with exactly one solution.
Answer: C. Solution: The first equation has slope \(-\frac{8}{13}\). A system has exactly one solution when the two lines have different slopes. Option C, \(-40x+13y=-830\), has slope \(\frac{40}{13}\), so it intersects the first line once. Detail Hint: Put each equation into slope-intercept form and choose the one with a different slope.
2.
Express each equation in slope-intercept form: \(-2x+4y=68\) and \(13x+4y=284\).
Answer: D. Solution: \(-2x+4y=68\Rightarrow 4y=2x+68\Rightarrow y=\frac{1}{2}x+17\). Also, \(13x+4y=284\Rightarrow 4y=-13x+284\Rightarrow y=-\frac{13}{4}x+71\). Detail Hint: Isolate \(y\) in each equation.
3.
Which linear system is represented by this graph?
Answer: A. Solution: The slanted line matches \(2x+5y=16\), and the vertical line is \(x=-1\). Detail Hint: A vertical line has equation \(x=\text{constant}\). Check which option includes \(x=-1\) and the correct slanted line.
4.
For each equation, identify a number you could multiply each term by so coefficients become integers. (1) \(\frac{5}{4}x+\frac{1}{6}y=\frac{47}{12}\). (2) \(\frac{6}{7}x-\frac{4}{5}y=16\).
Answer: B. Solution: For equation (1), the denominators are \(4,6,12\), so the LCD is \(12\). For equation (2), the denominators are \(7\) and \(5\), so the LCD is \(35\). Detail Hint: Use the least common denominator of all fractions in each equation.
5.
Which linear system is represented by this graph?
Answer: A. Solution: The rising line is \(x-y=3\), which becomes \(y=x-3\). The falling line is \(6x+5y=14\), which becomes \(y=-\frac{6}{5}x+\frac{14}{5}\). Detail Hint: Convert each option to slope-intercept form and compare slopes and intercepts with the graph.
6.
Create a linear system: An isosceles triangle has perimeter \(36\) cm. The base is \(9\) cm longer than each equal side.
Answer: C. Solution: Let \(s\) be each equal side and \(b\) be the base. The perimeter is \(2s+b=36\). Since the base is \(9\) cm longer than each equal side, \(s+9=b\). Detail Hint: An isosceles triangle has two equal sides.
7.
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) \(2l+2w=163;\ 2w=l-6\).
Answer: C. Solution: Situation A gives \(l=2w-6\), so it matches i. Situation B gives \(w=\frac{1}{2}(l-6)\), so it matches ii. Situation C gives \(l-6=2w\), so it matches iii. Detail Hint: Translate each sentence into an equation before matching.
8.
Create a linear system: Judy scored \(3\) points more than twice Ann's score. The total is \(39\) points.
Answer: B. Solution: Let \(j\) be Judy's score and \(a\) be Ann's score. Judy scored \(3\) more than twice Ann, so \(j=2a+3\). The total is \(j+a=39\). Detail Hint: Translate '3 more than twice Ann' as \(2a+3\).
9.
For what value of \(k\) does the linear system \(\frac{4}{5}x+y=14\) and \(kx+2y=28\) have infinite solutions?
Answer: C. Solution: Infinite solutions occur when both equations represent the same line. Multiply \(\frac{4}{5}x+y=14\) by \(2\): \(\frac{8}{5}x+2y=28\). Therefore, \(k=\frac{8}{5}\). Detail Hint: Make the second equation match a multiple of the first equation.
10.
Create a linear system: A rectangular field is \(35\) m longer than it is wide. The perimeter is \(290\) m.
Answer: B. Solution: Let \(l\) be length and \(w\) be width. Since the field is \(35\) m longer than wide, \(l=w+35\). The perimeter of a rectangle is \(2l+2w\), so \(2l+2w=290\). Detail Hint: Use one equation for the length-width relationship and one equation for perimeter.
11.
Create a linear system to model this situation: A length of outdoor lights is formed from strings that are \(5\) ft long and \(11\) ft long. Fourteen strings of lights are \(106\) ft long.
Answer: D. Solution: Let \(x\) be the number of \(5\)-ft strings and \(y\) be the number of \(11\)-ft strings. The number of strings gives \(x+y=14\). The total length gives \(5x+11y=106\). Detail Hint: One equation counts objects; the other equation counts total length.
12.
The solution of this linear system is \((-28,y)\). Determine \(y\). \(\frac{1}{2}x-\frac{1}{5}y=-\frac{86}{5}\); \(\frac{5}{6}x-4y=-\frac{262}{3}\).
Answer: A. Solution: Substitute \(x=-28\) into \(\frac{1}{2}x-\frac{1}{5}y=-\frac{86}{5}\): \(-14-\frac{1}{5}y=-\frac{86}{5}\). Multiply by \(5\): \(-70-y=-86\), so \(-y=-16\) and \(y=16\). Detail Hint: Since \(x\) is given, substitute directly into the first equation.
13.
Use the graph to approximate the solution of this linear system: \(6x-7y=-4\) and \(-\frac{3}{5}y=3x+7\).
Answer: D. Solution: The graph shows the two lines intersect at approximately \((-2.1,-1.2)\). Detail Hint: For graph questions, the solution of a linear system is the point where the two lines cross. Read the \(x\)-coordinate and \(y\)-coordinate of the intersection point.
14.
Which linear system has the solution \(x=8\) and \(y=2.5\)?
Answer: B. Solution: Substitute \(x=8\) and \(y=2.5\). Option B gives \(2(8)+2(2.5)=21\) and \(2(8)-2(2.5)=11\), so both equations are true. Detail Hint: Check the ordered pair in both equations of each system.
15.
Use substitution to solve this linear system: \(x-y=18\); \(\frac{3}{4}x+\frac{3}{4}y=-\frac{15}{2}\).
Answer: C. Solution: Multiply the second equation by \(4\): \(3x+3y=-30\). From \(x-y=18\), isolate \(x=y+18\). Substitute: \(3(y+18)+3y=-30\), so \(6y=-84\), \(y=-14\). Then \(x=-14+18=4\). Detail Hint: Clear fractions first, then substitute.
16.
Create a linear system to model this situation: Cheri charges \(\$19\) for a small lawn and \(\$29\) for a large lawn. She made \(\$287\) cutting \(13\) lawns.
Answer: B. Solution: Let \(s\) be the number of small lawns and \(l\) be the number of large lawns. The total number of lawns gives \(s+l=13\). The total money gives \(19s+29l=287\). Detail Hint: One equation counts lawns; the other equation counts dollars.
17.
Which of the following best describes the lines \(y=2x-2\) and \(y=2x+7\)?
Answer: C. Solution: Both lines have slope \(2\), but their \(y\)-intercepts are different: \(-2\) and \(7\). Same slope and different intercepts means the lines are parallel. Detail Hint: Compare the slopes and \(y\)-intercepts.
18.
Create a linear system to model this situation: Tickets for a school play cost \(\$8\) for adults and \(\$4.75\) for students. There were ten more student tickets sold than adult tickets, and a total of \(\$1399\) in ticket sales was collected.
Answer: B. Solution: Let \(a\) be adult tickets and \(s\) be student tickets. Ten more student tickets than adult tickets gives \(s=a+10\). The revenue equation is \(8a+4.75s=1399\). Detail Hint: Use ticket prices for the money equation and the phrase 'ten more students' for the second equation.
19.
Use substitution to solve this problem: Wai Sen scored \(85\%\) on part A and \(95\%\) on part B. Her total mark was \(70\), and the total possible was \(78\). How many marks is each part worth?
Answer: A. Solution: Let \(a\) be part A and \(b\) be part B. Then \(a+b=78\) and \(0.85a+0.95b=70\). Substitute \(a=78-b\): \(0.85(78-b)+0.95b=70\). Then \(66.3-0.85b+0.95b=70\), so \(0.1b=3.7\), \(b=37\). Then \(a=41\). Detail Hint: Use one equation for total possible marks and one equation for weighted score.
20.
Create a linear system to model this situation: A woman is \(3\) times as old as her son. In thirteen years, she will be \(2\) times as old as her son will be.
Answer: C. Solution: Let \(w\) be the woman's age and \(s\) be the son's age. Currently, \(w=3s\). In thirteen years, the woman will be \(w+13\) and the son will be \(s+13\), so \(w+13=2(s+13)\). Detail Hint: Add \(13\) to both ages before writing the future-age equation.
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