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To determine if the graphs of the equations [tex]\( 2x - 3y = 1 \)[/tex] and [tex]\( 2x + 3y = 2 \)[/tex] are intersecting lines using addition (also known as the elimination method), we first add the two equations together. By adding [tex]\( 2x - 3y = 1 \)[/tex] and [tex]\( 2x + 3y = 2 \)[/tex], we get:
[tex]\[ (2x - 3y) + (2x + 3y) = 1 + 2 \][/tex]
This simplifies to:
[tex]\[ 4x = 3 \][/tex]
Solving for [tex]\( x \)[/tex] gives us:
[tex]\[ x = \frac{3}{4} \][/tex]
Next, we substitute [tex]\( x = \frac{3}{4} \)[/tex] back into one of the original equations to solve for [tex]\( y \)[/tex]. Using the equation [tex]\( 2x - 3y = 1 \)[/tex], we get:
[tex]\[ 2\left(\frac{3}{4}\right) - 3y = 1 \][/tex]
Simplifying, we obtain:
[tex]\[ \frac{3}{2} - 3y = 1 \][/tex]
Subtracting [tex]\(\frac{3}{2}\)[/tex] from both sides gives:
[tex]\[ -3y = 1 - \frac{3}{2} = -\frac{1}{2} \][/tex]
Solving for [tex]\( y \)[/tex], we get:
[tex]\[ y = \frac{1}{6} \][/tex]
Thus, the solution to the system is [tex]\( x = \frac{3}{4} \)[/tex] and [tex]\( y = \frac{1}{6} \)[/tex], indicating that the lines intersect at the point [tex]\(\left(\frac{3}{4}, \frac{1}{6}\right)\)[/tex]. This confirms that the lines do indeed intersect at this point.
[tex]\[ (2x - 3y) + (2x + 3y) = 1 + 2 \][/tex]
This simplifies to:
[tex]\[ 4x = 3 \][/tex]
Solving for [tex]\( x \)[/tex] gives us:
[tex]\[ x = \frac{3}{4} \][/tex]
Next, we substitute [tex]\( x = \frac{3}{4} \)[/tex] back into one of the original equations to solve for [tex]\( y \)[/tex]. Using the equation [tex]\( 2x - 3y = 1 \)[/tex], we get:
[tex]\[ 2\left(\frac{3}{4}\right) - 3y = 1 \][/tex]
Simplifying, we obtain:
[tex]\[ \frac{3}{2} - 3y = 1 \][/tex]
Subtracting [tex]\(\frac{3}{2}\)[/tex] from both sides gives:
[tex]\[ -3y = 1 - \frac{3}{2} = -\frac{1}{2} \][/tex]
Solving for [tex]\( y \)[/tex], we get:
[tex]\[ y = \frac{1}{6} \][/tex]
Thus, the solution to the system is [tex]\( x = \frac{3}{4} \)[/tex] and [tex]\( y = \frac{1}{6} \)[/tex], indicating that the lines intersect at the point [tex]\(\left(\frac{3}{4}, \frac{1}{6}\right)\)[/tex]. This confirms that the lines do indeed intersect at this point.
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