Houghton Mifflin Harcourt Algebra 1, 2015
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Houghton Mifflin Harcourt Algebra 1, 2015 View details
Unit 5 Assessment Readiness

Exercise 4 Page 497

a

Let's call the number of small cabinets s, and the number of large cabinets l. We know that Asif bought a total of 13 cabinets which we can express with the following equation.

s+l = 13 Additionally, we know that a small cabinet cost $43.50 and a large cabinet cost $65.95. By multiplying these costs with the number of each cabinet you bought, we should get a total cost of $745.10 according to the exercise. 43.50s+65.95l = 745.10 If we combine these equations, we get a system of equations. s + l = 13 & (I) 43.50s + 65.95l = 745.10 & (II) Since s and l have coefficients of 1 in the first equation, the easiest way to solve these equations is to solve for one of these variables and then use the Substitution Method.

s + l = 13 & (I) 43.50s + 65.95l = 745.10 & (II)
s=13-l 43.50s + 65.95l = 745.10
s=13-l 43.50( 13-l) + 65.95l = 745.10
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(II): Solve for l
s=13-l 565.50-43.50l + 65.95l = 745.10
s=13-l 565.50+22.45l = 745.10
s=13-l 22.45l = 179.60
s=13-l l = 8

Now that we have solved for the l-variable, we can substitute that value into the first equation and solve for the s-variable.

s=13-l l = 8
s=13- 8 l = 8
s=5 l = 8

Asif bought 5 small cabinets and 8 large cabinets. The statement is therefore true.

b

From Part A we know that Asif bough 5 small cabinets and 8 large cabinets. Therefore, the statement that he bought 7 cabinets, is false.

c

From Part A we know that Asif bough 8 large cabinets. If we multiply this number by the cost of a large cabinet, we get the total expenditures on this type of cabinet.

69.95(8)=559.6 The total cost for the large cabinets was $559.6. Therefore, the statement is false.