[tex] \huge \boxed{\mathfrak{Question} \downarrow}[/tex]
Solve the following using Substitution method
2x – 5y = -13
3x + 4y = 15
[tex] \large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}[/tex]
[tex]\left. \begin{array} { l } { 2 x - 5 y = - 13 } \\ { 3 x + 4 y = 15 } \end{array} \right.[/tex]
- To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.
[tex]2x-5y=-13, \: 3x+4y=15 [/tex]
- Choose one of the equations and solve it for x by isolating x on the left-hand side of the equal sign. I'm choosing the 1st equation for now.
[tex]2x-5y=-13 [/tex]
- Add 5y to both sides of the equation.
[tex]2x=5y-13 [/tex]
[tex]x=\frac{1}{2}\left(5y-13\right) \\ [/tex]
- Multiply [tex]\frac{1}{2}\\[/tex] times 5y - 13.
[tex]x=\frac{5}{2}y-\frac{13}{2} \\ [/tex]
- Substitute [tex]\frac{5y-13}{2}\\[/tex] for x in the other equation, 3x + 4y = 15.
[tex]3\left(\frac{5}{2}y-\frac{13}{2}\right)+4y=15 \\ [/tex]
- Multiply 3 times [tex]\frac{5y-13}{2}\\[/tex].
[tex]\frac{15}{2}y-\frac{39}{2}+4y=15 \\ [/tex]
- Add [tex]\frac{15y}{2} \\[/tex] to 4y.
[tex]\frac{23}{2}y-\frac{39}{2}=15 \\ [/tex]
- Add [tex]\frac{39}{2}\\[/tex] to both sides of the equation.
[tex]\frac{23}{2}y=\frac{69}{2} \\ [/tex]
- Divide both sides of the equation by 23/2, which is the same as multiplying both sides by the reciprocal of the fraction.
[tex]\large \underline{ \underline{ \sf \: y=3 }}[/tex]
- Substitute 3 for y in [tex]x=\frac{5}{2}y-\frac{13}{2}\\[/tex]. Because the resulting equation contains only one variable, you can solve for x directly.
[tex]x=\frac{5}{2}\times 3-\frac{13}{2} \\ [/tex]
[tex]x=\frac{15-13}{2} \\ [/tex]
- Add [tex]-\frac{13}{2}\\[/tex] to [tex]\frac{15}{2}\\[/tex] by finding a common denominator and adding the numerators. Then reduce the fraction to its lowest terms if possible.
[tex] \large\underline{ \underline{ \sf \: x=1 }}[/tex]
- The system is now solved. The value of x & y will be 1 & 3 respectively.
[tex] \huge\boxed{ \boxed{\bf \: x=1, \: y=3 }}[/tex]
