PROBLEM_1. Tickets cost $localeToFixed(A1, 2) each for adults and $localeToFixed(B1, 2) each for kids, and the group paid $localeToFixed(C1, 2) in total. There were abs(C2) fewer adults than kids in the group.
PROBLEM_2.
C1 people attended a baseball game. Everyone there was a fan of either the home team or the away team.
The number of home team fans was abs(C2) less than abs(B2) times the number of away team fans.
How many home team and away team fans attended the game?
# of UNIT_1 = X
# of UNIT_2 = Y
Let x equal the number of UNIT_1 and y equal the number of UNIT_2.
The system of equations is then:
\blue{expr(["+", ["*", A1, "x"], ["*", B1, "y"]]) = C1}
\green{x = expr(["+", ["*", -B2, "y"], C2])}
Solve for x and y using substitution.
Since x has already been solved for, substitute \green{expr(["+", ["*", -B2, "y"], C2])} for x in the first equation.
\blue{A1-}\green{(expr(["+", ["*", -B2, "y"], C2]))}\blue{+ expr(["*", B1, "y"]) = C1}
Simplify and solve for y.
expr(["+", ["*", roundTo(8, A1 * -B2), "y"], roundTo(8, A1 * C2)]) + expr(["*", B1, "y"]) = C1
expr(["+", ["*", roundTo(8, A1 * -B2 + B1), "y"], roundTo(8, A1 * C2)]) = C1
expr(["*", roundTo(8, A1 * -B2 + B1), "y"]) = roundTo(8, C1 - A1 * C2)
y = \dfrac{roundTo(8, C1 - A1 * C2)}{roundTo( 8, A1 * -B2 + B1 )}
\orange{y = Y}
Now that you know \orange{y = Y}, plug it back into \green{x = expr(["+", ["*", -B2, "y"], C2])} to find x.
\green{x = -B2-}\orange{(Y)}\green{ + C2}
x = -B2 * Y + C2
\red{x = X}
You can also plug \orange{y = Y} into \blue{expr(["+", ["*", A1, "x"], ["*", B1, "y"]]) = C1} and get the same answer for x:
\blue{expr(["*", A1, "x"]) + B1-}\orange{(Y)}\blue{= C1}
\red{x = X}
There were X UNIT_1 and Y UNIT_2.
PROBLEM_1. Bags of candy cost $localeToFixed(A1, 2), and bags of cookies cost $localeToFixed(B1, 2), and sales equaled $localeToFixed(C1, 2) in total. There were C2 more bags of cookies than candy sold.
PROBLEM_2.
The sum of two angles is C1^\circ. Angle 2 is abs(C2)^\circ smaller than abs(A2) times angle 1.
What are the measures of the two angles in degrees?
# of UNIT_1 = X
# of UNIT_2 = Y
Let x equal the number of UNIT_1 and y equal the number of UNIT_2.
Let x equal the measure of UNIT_1 and y equal the measure of UNIT_2.
The system of equations is then:
\blue{expr(["+", ["*", A1, "x"], ["*", B1, "y"]]) = C1}
\green{y = expr(["+", ["*", -A2, "x"], C2])}
Since we already have solved for y in terms of x, we can use substitution to solve for x and y.
Substitute \green{expr(["+", ["*", -A2, "x"], C2])} for y in the first equation.
\blue{expr(["*", A1, "x"]) + B1-}\green{(expr(["+", ["*", -A2, "x"], C2]))}\blue{= C1}
Simplify and solve for x.
expr(["+", ["*", A1, "x"], ["*", roundTo(8, B1 * -A2), "x"]]) + roundTo(8, B1 * C2) = C1
expr(["+", ["*", roundTo(8, A1 + B1 * -A2), "x"], roundTo(8, B1 * C2)]) = C1
expr(["*", roundTo(8, A1 + B1 * -A2), "x"]) = roundTo(8, C1 - B1 * C2)
x = \dfrac{roundTo( 8, C1 - B1 * C2 )}{roundTo(8, A1 + B1 * -A2)}
\red{x = X}
Now that you know \red{x = X}, plug it back into \green{y = expr(["+", ["*", -A2, "x"], C2])} to find y.
\green{y = -A2-}\red{(X)}\green{ + C2}
y = roundTo(8, -A2 * X) + C2
\orange{y = Y}
You can also plug \red{x = X} into \blue{expr(["+", ["*", A1, "x"], ["*", B1, "y"]]) = C1} and get the same answer for y:
\blue{A1-}\red{(X)}\blue{ + expr(["*", B1, "y"]) = C1}
\orange{y = Y}
X bags of candy and Y bags of cookies were sold.
The measure of angle 1 is X^\circ and the measure of angle 2 is Y^\circ.
All of the {3rd|4th|5th} grade teachers and students from school(1) went on a field trip to an {art|archaeology} museum. Tickets were $localeToFixed(A1, 2) each for teachers and $localeToFixed(B1, 2) each for students, and the group paid $localeToFixed(C1, 2) in total.
{A few weeks later|The next month}, the same group visited a {science|natural history} museum where the tickets cost $localeToFixed(A2, 2) each for teachers and $localeToFixed(B2, 2) each for students, and the group paid $localeToFixed(C2, 2) in total.
Find the number of teachers and students on the field trips.
# of teachers = X
# of students = Y
Let x equal the number of teachers and y equal the number of students.
The system of equations is:
\blue{expr(["+", ["*", A1, "x"], ["*", B1, "y"]]) = C1}
\green{expr(["+", ["*", A2, "x"], ["*", B2, "y"]]) = C2}
Solve for x and y using elimination.
Multiply the bottom equation by MULT2.
Multiply the top equation by MULT1 and the bottom equation by MULT2.
Multiply the top equation by MULT1.
\blue{expr(["+", ["*", A1 * MULT1, "x"], ["*", B1 * MULT1, "y"]]) = C1 * MULT1}
\green{expr(["+", ["*", A2 * MULT2, "x"], ["*", B2 * MULT2, "y"]]) = C2 * MULT2}
Add the top and bottom equations together.
expr(["*", roundTo(8, B1 * MULT1 + B2 * MULT2), "y"]) = roundTo(8, C1 * MULT1 + C2 * MULT2)
y = \dfrac{roundTo(8, C1 * MULT1 + C2 * MULT2)}{roundTo(8, B1 * MULT1 + B2 * MULT2)}
\orange{y = Y}
Now that you know \orange{y = Y}, plug it back into \blue{expr(["+", ["*", A1, "x"], ["*", B1, "y"]]) = C1} to find x.
\blue{expr(["*", A1, "x"]) + B1-}\orange{(Y)}\blue{= C1}
expr(["+", ["*", A1, "x"], B1 * Y]) = C1
expr(["*", A1, "x"]) = roundTo(8, C1 - B1 * Y)
x = \dfrac{roundTo( 8, C1 - B1 * Y )}{A1}
\red{x = X}
You can also plug \orange{y = Y} into \green{expr(["+", ["*", A2, "x"], ["*", B2, "y"]]) = C2} and get the same answer for x:
\green{expr(["*", A2, "x"]) + B2-}\orange{(Y)}\green{= C2}
\red{x = X}
There were X teachers and Y students on the field trips.
The sum of two numbers is C1, and their difference is C2. What are the two numbers?
larger number = LARGER
smaller number = SMALLER
Let x be the first number, and let y be the second number.
The system of equations is:
\blue{expr(["+", ["*", A1, "x"], ["*", B1, "y"]]) = C1}
\green{expr(["+", ["*", A2, "x"], ["*", B2, "y"]]) = C2}
Solve for x and y using elimination.
Multiply the bottom equation by MULT2.
Multiply the top equation by MULT1 and the bottom equation by MULT2.
Multiply the top equation by MULT1.
\blue{expr(["+", ["*", A1 * MULT1, "x"], ["*", B1 * MULT1, "y"]]) = C1 * MULT1}
\green{expr(["+", ["*", A2 * MULT2, "x"], ["*", B2 * MULT2, "y"]]) = C2 * MULT2}
Add the top and bottom equations together.
expr(["*", roundTo(8, A1 * MULT1 + A2 * MULT2), "x"]) = roundTo(8, C1 * MULT1 + C2 * MULT2)
x = \dfrac{roundTo(8, C1 * MULT1 + C2 * MULT2)}{roundTo( 8, A1 * MULT1 + A2 * MULT2 )}
\red{x = X}
Now that you know \red{x = X}, plug it back into \blue{expr(["+", ["*", A1, "x"], ["*", B1, "y"]]) = C1} to find y.
\blue{A1-}\red{(X)}\blue{ + expr(["*", B1, "y"]) = C1}
expr(["+", A1 * X, ["*", B1, "y"]]) = C1
expr(["*", B1, "y"]) = roundTo( 8, C1 - A1 * X )
\dfrac{expr(["*", B1, "y"])}{\blue{B1}} = \dfrac{roundTo( 8, C1 - A1 * X )}{\blue{B1}}
\orange{y = Y}
You can also plug \red{x = X} into \green{expr(["+", ["*", A2, "x"], ["*", B2, "y"]]) = C2} and get the same answer for y:
\green{A2-}\red{(X)}\green{ + expr(["*", B2, "y"]) = C2}
\orange{y = Y}
Therefore, the larger number is LARGER, and the smaller number is SMALLER.