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Algebra Test: Linear Equations, Graphs, and Systems of Equations - Document preview page 1

Algebra Test: Linear Equations, Graphs, and Systems of Equations - Page 1

Document preview content for Algebra Test: Linear Equations, Graphs, and Systems of Equations

Algebra Test: Linear Equations, Graphs, and Systems of Equations

A test focusing on solving linear equations, graphing equations, and solving systems of equations in algebra.

Hunter Harris
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Algebra Test: Linear Equations, Graphs, and Systems of Equations - Page 1 preview imageAlgebra Test: Linear Equations, Graphs, and Systems of EquationsClass: Algebra 1 Part 2U1 MidReport Date: 02/26/2013PrintTest Questions1.Find the slope of the line.A.1B.0C.2D.PointsPossible:0.472.Find thex-andy-intercept of the line.2x+ 3y=–18A.x-intercept is 18;y-intercept is 18.B.x-intercept is–6;y-intercept is–9.C.x-intercept is 2;y-intercept is 3.D.x-intercept is–9;y-intercept is–6.PointsPossible:0.473.Find the slope andy-intercept of the line.
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Algebra Test: Linear Equations, Graphs, and Systems of Equations - Page 2 preview image
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Algebra Test: Linear Equations, Graphs, and Systems of Equations - Page 3 preview imagey=x+ 1A.;–1B.–1;C.; 1D.1;PointsPossible:0.474.Write an equation of a line with thegiven slope andy-intercept.m=,b=A.y=x–B.y=x–C.y=x+D.y=x+PointsPossible:0.475.Use the slope andy-intercept to graph the equation.y=x+4A.
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Algebra Test: Linear Equations, Graphs, and Systems of Equations - Page 4 preview imageB.C.D.
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Algebra Test: Linear Equations, Graphs, and Systems of Equations - Page 5 preview imagePointsPossible:0.476.Match the equation with its graph.–6x–4y= 24A.B.C.
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Algebra Test: Linear Equations, Graphs, and Systems of Equations - Page 6 preview imageD.PointsPossible:0.477.A line passes through (3,–2) and (8, 2).a.Write an equation for the line in point-slope form.b.Rewrite the equation in standard form using integers.A.y+ 2 =(x+ 3);–4x+ 5y=–22B.y–3 =(x+ 2);–4x+ 5y= 23C.y–2 =(x–3);–4x+ 5y= 22D.y+ 2 =(x–3);–4x+ 5y=–22PointsPossible:0.478.The rate of change is constant in each table. Find the rate of change. Explainwhat the rate of change means for the situation.
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Algebra Test: Linear Equations, Graphs, and Systems of Equations - Page 7 preview imageTime (days)Cost ($)375410051256150A.dollars per day; the cost is $25 for eachday.B.dollars per day; the cost is $25 for eachday.C.dollars per day; the cost is $75 for eachday.D.dollars per day; the costs $1 for 150daysPointsPossible:0.479.State whether the slope is 0 or undefined.A.0B.undefinedPointsPossible:0.4710.State whether the slope is 0 or undefined.
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Algebra Test: Linear Equations, Graphs, and Systems of Equations - Page 8 preview imageA.undefinedB.0PointsPossible:0.4711.Use the graph.a.Which plant was the tallest at the beginning?b.Which plant had the greatest rate of change over the 6 weeks?A.plant 2; plant 2B.plant 1; plant 3C.plant 3; plant 1
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Algebra Test: Linear Equations, Graphs, and Systems of Equations - Page 9 preview imageD.plant 3; plant 3PointsPossible:0.4712.Writey=x +9in standard form using integers.A.–x–4y= 36B.–x+ 4y= 9C.4x–y= 36D.–x+ 4y= 36PointsPossible:0.4713.Writey=x +3 in standard form using integers.A.–2x+ 3y= 9B.–2x+ 3y= 3C.3x–2y= 9D.–2x–3y= 9PointsPossible:0.4714.Is the relationship shown by the data linear? If so, model the data with anequation.xy329715122117A.The relationship is linear;y–2 =(x–3).B.The relationship is linear;y–3 =(x–2).C.The relationship is linear;y–2 =(x–3).D.The relationship is not linear.PointsPossible:0.4715.
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Algebra Test: Linear Equations, Graphs, and Systems of Equations - Page 10 preview imageThe table shows the height of a plant as it grows.a.Model the data with an equation.b.Based on your model, predict the height of the plant at 12 months.Time(months)Plant Height(cm)324540756972A.y–24 =(x–3); 48 cmB.y–24 = 8(x–3); 96 cmC.y–3 =(x–24); 84 cmD.The relationship cannot be modeled.PointsPossible:0.4716.Find the slope of the line.A.B.C.D.PointsPossible:0.4717.Write an equation of a line with the given slope andy-intercept.m= 1,b= 4
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Algebra Test: Linear Equations, Graphs, and Systems of Equations - Page 11 preview imageA.y= 4x+ 1B.y=x–4C.y=–1x+ 4D.y=x+ 4PointsPossible:0.4718.Find the slope of the line that passes through the pair of points.(1, 7), (10, 1)A.B.C.D.PointsPossible:0.4719.Giselle pays $240 in advance on her account at the athletic club. Each time she usesthe club, $15 is deducted from the account. Model the situation with a linear functionand a graph.A.b= 240–15xB.
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Algebra Test: Linear Equations, Graphs, and Systems of Equations - Page 12 preview imageb= 240 + 15xC.b= 225 + 15xD.b= 225–15xPointsPossible:0.9420.Find thex-andy-intercept of the line.
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Algebra Test: Linear Equations, Graphs, and Systems of Equations - Page 13 preview image–3x–2y= 12A.x-intercept is–6;y-intercept is–4.B.x-intercept is–4;y-intercept is–6.C.x-intercept is–2; y-intercept is–3.D.x-intercept is–3;y-intercept is–2.PointsPossible:0.4721.The grocery store sells kumquats for $4.75 a pound andAsian pears for $3.75 apound. Write an equation in standard form for the weights of kumquatskand Asianpearspthat a customer could buy with $12.A.4.75k =3.75p +12B.4.75k +3.75p =12C.4.75p +3.75k =12D.4.75 + 3.75 =kPointsPossible:0.4722.The grocery store sells kumquats for $4.50 a pound and Asian pears for $2.50 apound. Write an equation in standard form for the weights of kumquatskand Asianpearspthat acustomer could buy with $12.A.4.5 + 2.5 =kB.4.5p +2.5k =12C.4.5k =2.5p +12D.4.5k +2.5p =12PointsPossible:0.4723.Write an equation in point-slope form for the linethrough the given pointwith the given slope.(4, 3);m=A.y+ 3 =(x+ 4)B.y–4 =(x–3)C.y+ 3 =(x–4)D.y–3 =(x–4)
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Algebra Test: Linear Equations, Graphs, and Systems of Equations - Page 14 preview imagePointsPossible:0.4724.A line passesthrough (–10, 3) and (–7, 10).a.Write an equation for the line in point-slope form.b.Rewrite the equation in standard form using integers.A.y–3 =(x–10);–7x+ 3y= 79B.y+ 3 =(x+ 10);–7x+ 3y=–79C.y–3 =(x+ 10);–7x+ 3y= 79D.y+ 10 =(x–3);–7x+ 3y=–51PointsPossible:0.4725.The table shows the height of a plant as it grows.a.Model the data with an equation.b.Based on your model, predict theheight of the plant at 12 months.Time(months)Plant Height(cm)39515721927A.y–3 =(x–9); 39 cmB.y–9 = 3(x–3); 36 cmC.y–9 =(x–3); 18 cmD.Therelationship cannot be modeled.PointsPossible:0.4726.Find the slope andy-intercept of the line.18x+ 2y=–36A.18;–36B.;C.
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Algebra Test: Linear Equations, Graphs, and Systems of Equations - Page 15 preview image;D.;PointsPossible:0.4727.The grocery store sells kumquats for $3.75 a pound and Asian pears for $2.25 apound. Write an equation in standard form for the weights of kumquatskand Asianpearspthat a customer could buy with $22.A.3.75p +2.25k =22B.3.75k +2.25p =22C.3.75 + 2.25 =kD.3.75k =2.25p +22PointsPossible:0.4728.A line passes through (–5,–3) and (–3,–2).a.Write an equation for the line inpoint-slope form.b.Rewrite the equation in standard form using integers.A.y–3 =(x+ 5);–x+ 2y= 1B.y+ 3 =(x–5);–x+ 2y=–1C.y+ 5 =(x+ 3);–x+ 2y=–7D.y+ 3 =(x+ 5);–x+ 2y=–1PointsPossible:0.4729.In April, you have a balance of $75 in your bank account. Each month you deposit$25. Let January=1, February=2, and so on. Write an equation for this situation.Use the equation to find thebalance in June.A.y= 25(x–4); $50B.y–75 = 25(x–4) ; $125C.y–75 = 25x; $25D.y= 25(x–4); $75PointsPossible:0.4730.Is the relationship shown by the datalinear? If so, model the data with anequation.
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Algebra Test: Linear Equations, Graphs, and Systems of Equations - Page 16 preview imagexy9–512–615–718–8A.The relationship is not linear.B.The relationship is linear;y+ 5 =(x–9).C.The relationship is linear;y–9 =(x+ 5).D.The relationship is linear;y+ 5 =(x–9).PointsPossible:0.4731.The table shows the height of a plant as it grows.a.Model the data with an equation.b.Based on your model, predict the height of theplant at 12 months.Time(months)Plant Height(cm)318530742954A.y–18 =(x–3); 36 cmB.y–18 = 6(x–3); 72 cmC.The relationship cannot be modeled.D.y–3 =(x–18); 66 cmPointsPossible:0.4732.A line passes through (6,–9) and (10,–2).a.Write an equation for the line in point-slope form.b.Rewrite the equation in standard form using integers.A.y–9 =(x–6);–7x+ 4y=78B.y–6 =(x+ 9);–7x+ 4y= 87C.
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