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matt did an experiment to study the solubility of two substances. he poured 100 ml of water at 20 °c into each of two beakers labeled a and b. he put 50 g of substance a in the beaker labeled a and 50 g of substance b in the beaker labeled b. the solution in both beakers was stirred for 1 minute. the amount of substance left undissolved in the beakers was weighed. the experiment was repeated for different temperatures of water and the observations were recorded as shown.

part 1: which, if any, substance is soluble in water?
part 2: explain how the data you determine solubility for both substances for temperatures 20 °c to 80 °c.

step-by-step explanation:

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3