ExpandStr\( \mathrm { 2 \cdot x - 5 = y } \)ProcessLogAdd \( \mathrm { 5 } \) to both sidesStr\( \mathrm { 2 \cdot x - 5 + 5 = y + 5 } \)ProcessLog\( \mathrm { -5 + 5 = 0 } \)Str\( \mathrm { 2 \cdot x + 0 = y + 5 } \)ProcessLogMultiply both sides with \( \mathrm { \displaystyle\frac{1}{2} } \)Str\( \mathrm { \displaystyle\frac{2}{2} \cdot x = \left(y + 5\right) \cdot \displaystyle\frac{1}{2} } \)StrProcessLog\( \mathrm { \displaystyle\frac{2}{2} \cdot x = \left(y + 5\right) \cdot \displaystyle\frac{1}{2} } \) \( \mathrm { \Longrightarrow } \) \( \mathrm { x = \displaystyle\frac{5}{2} + \displaystyle\frac{y}{2} } \)Str\( \mathrm { x = \displaystyle\frac{5}{2} + \displaystyle\frac{y}{2} } \)\( \mathrm { y = 2 \cdot x - 5 } \)\( \mathrm { x = \displaystyle\frac{5}{2} + \displaystyle\frac{y}{2} } \)Isolate \( \mathrm { x } \)Str\( \mathrm { y = - x + 7 } \)ProcessLogReplace \( \mathrm { x } \) with \( \mathrm { \displaystyle\frac{5}{2} + \displaystyle\frac{y}{2} } \)Str\( \mathrm { y = - (\displaystyle\frac{5}{2} + \displaystyle\frac{y}{2}) + 7 } \)ExpandProcessLogSet right side of the relation to zeroStr\( \mathrm { y - (- (\displaystyle\frac{5}{2} + \displaystyle\frac{y}{2}) + 7) = 0 } \)Str\( \mathrm { y - (- (\displaystyle\frac{5}{2} + \displaystyle\frac{y}{2}) + 7) } \)ProcessLogUse the distributive property of multiplication: \n \( \mathrm { a \cdot ( b + c ) = a \cdot b + a \cdot c } \) \n \( \mathrm { - (- (\displaystyle\frac{5}{2} + \displaystyle\frac{y}{2}) + 7) = -7 + \displaystyle\frac{5}{2} + \displaystyle\frac{y}{2} } \)Str\( \mathrm { y - 7 + \displaystyle\frac{5}{2} + \displaystyle\frac{y}{2} } \)ProcessLog\( \mathrm { -7 + \displaystyle\frac{5}{2} = - \displaystyle\frac{9}{2} } \)Str\( \mathrm { y - \displaystyle\frac{9}{2} + \displaystyle\frac{y}{2} = 0 } \)ProcessLog\( \mathrm { y + \displaystyle\frac{y}{2} = \displaystyle\frac{3 \cdot y}{2} } \)Str\( \mathrm { \displaystyle\frac{3 \cdot y}{2} - \displaystyle\frac{9}{2} = 0 } \)StrStrStr\( \mathrm { \displaystyle\frac{3 \cdot y}{2} - \displaystyle\frac{9}{2} } \)ProcessLogRewrite \( \mathrm { \displaystyle\frac{3 \cdot y}{2} - \displaystyle\frac{9}{2} } \) with using the common denominator \( \mathrm { 2 } \) : \n \( \mathrm { \displaystyle\frac{3 \cdot y - 9}{2} } \)Str\( \mathrm { \displaystyle\frac{3 \cdot y - 9}{2} } \)ExpandStr\( \mathrm { 3 \cdot y - 9 = 0 } \)ProcessLogAdd \( \mathrm { 9 } \) to both sidesStr\( \mathrm { 3 \cdot y - 9 + 9 = 0 + 9 } \)ProcessLog\( \mathrm { -9 + 9 = 0 } \)Str\( \mathrm { 3 \cdot y + 0 = 0 + 9 } \)ProcessLogMultiply both sides with \( \mathrm { \displaystyle\frac{1}{3} } \)Str\( \mathrm { \displaystyle\frac{3}{3} \cdot y = 9 \cdot \displaystyle\frac{1}{3} } \)StrProcessLog\( \mathrm { \displaystyle\frac{3}{3} \cdot y = 9 \cdot \displaystyle\frac{1}{3} } \) \( \mathrm { \Longrightarrow } \) \( \mathrm { y = 3 } \)Str\( \mathrm { y = 3 } \)\( \mathrm { 3 \cdot y - 9 = 0 } \)\( \mathrm { y = 3 } \)Solve for \( \mathrm { 3 \cdot y - 9 = 0 } \)\( \mathrm { y = 3 } \)Isolate \( \mathrm { y } \)Str\( \mathrm { x = \displaystyle\frac{5}{2} + \displaystyle\frac{y}{2} } \)ProcessLogReplace \( \mathrm { y } \) with \( \mathrm { 3 } \)Str\( \mathrm { x = \displaystyle\frac{5}{2} + \displaystyle\frac{3}{2} } \)ExpandProcessLog\( \mathrm { \displaystyle\frac{5}{2} + \displaystyle\frac{3}{2} = 4 } \)Str\( \mathrm { x = 4 } \)\( \mathrm { x = 4 } \)Simplify

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Solution: Find the point of intersection of y equals 2 x minus 5 and y equals negative x plus 7.

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