randRangeNonZero( -20, 20 ) randRange( 2, 8 ) randRangeNonZero( -200, 200 ) randRange( 2, 9 )
180 - (A + C) * SOLUTION - B A*SOLUTION + B shuffle( randFromArray( [ [ 3, 4 ], [ 2, 5 ] ] ) ) $._("vertical angles") $._("alternate interior angle")

The two horizontal lines are parallel, and there is a third line that intersects them as shown below.

Solve for x:

SOLUTION

var eq1 = A + "x + " + B + "^\\circ"; var eq2 = C + "x + " + D + "^\\circ"; init({ range: [ [ -1, 11 ], [-1, 4] ] }); graph.pl = new ParallelLines( 0, 0, 10, 0, 3 ); graph.pl.draw(); graph.pl.drawMarkers(ANCHOR); graph.pl.drawTransverse( ANCHOR ); graph.pl.drawAngle( KNOWN_INDEX, eq1 ); graph.pl.drawAngle( UNKNOWN_INDEX, eq2, "#FFA500" );

The pink angles are adjacent to the blue angle and form a straight line, so we know that:

\color{BLUE}{Ax + B} + \color{PINK}{y} = 180

The pink angles equal each other because they are \color{GREEN}{\text{VERTICAL_ANGLES}}.

graph.pl.drawAdjacentAngles( KNOWN_INDEX, "y^\\circ", "#FF00A5" );

One of the pink angles \color{GREEN}{corresponds} with the orange angle, and the other pink angle forms an \color{GREEN}{\text{ALTERNATE_INTERIOR_ANGLE}}. Therefore, the orange angle measure equals the pink angle measure.

\color{PINK}{y} = \color{ORANGE}{Cx + D}

Substitute \color{ORANGE}{Cx + D} for \color{PINK}{y} in our first equation.

\color{BLUE}{Ax + B} + \color{ORANGE}{Cx + D} = 180

Combine like terms.

A + Cx + B + D = 180

Subtract \color{PINK}{abs(B + D)} from both sides.

Add \color{PINK}{abs(B + D)} to both sides.

(A + Cx + B + D) \color{PINK}{+ -(B + D)} = 180 \color{PINK}{+ -(B + D)}

A + Cx = 180 - B - D

Divide by \color{PINK}{A + C}.

\dfrac{A + Cx}{\color{PINK}{A + C}} = \dfrac{180 - B - D}{\color{PINK}{A + C}}

Simplify.

x = (180 - B - D) / (A + C)

Note that the blue and orange angles are \color{GREEN}{supplementary}.