Question

Solve the following using Substitution method
2x – 5y = -13

3x + 4y = 15

Answer 1

`\huge \boxed{\mathfrak{Question} \downarrow}`

Solve the following using Substitution method

2x – 5y = -13

3x + 4y = 15

`\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}`

`\left. \begin{array} { l } { 2 x - 5 y = - 13 } \\ { 3 x + 4 y = 15 } \end{array} \right.`

• 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.

`2x-5y=-13, \: 3x+4y=15`

• 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.

`2x-5y=-13`

• Add 5y to both sides of the equation.

`2x=5y-13`

• Divide both sides by 2.

`x=(1)/(2)\left(5y-13\right) \\`

• Multiply
  `(1)/(2)\\` times 5y - 13.

`x=(5)/(2)y-(13)/(2) \\`

• Substitute
  `(5y-13)/(2)\\` for x in the other equation, 3x + 4y = 15.

`3\left((5)/(2)y-(13)/(2)\right)+4y=15 \\`

• Multiply 3 times
  `(5y-13)/(2)\\`.

`(15)/(2)y-(39)/(2)+4y=15 \\`

• Add
  `(15y)/(2) \\` to 4y.

`(23)/(2)y-(39)/(2)=15 \\`

• Add
  `(39)/(2)\\` to both sides of the equation.

`(23)/(2)y=(69)/(2) \\`

• Divide both sides of the equation by 23/2, which is the same as multiplying both sides by the reciprocal of the fraction.

`\large \underline{ \underline{ \sf \: y=3 }}`

• Substitute 3 for y in
  `x=(5)/(2)y-(13)/(2)\\`. Because the resulting equation contains only one variable, you can solve for x directly.

`x=(5)/(2)* 3-(13)/(2) \\`

• Multiply 5/2 times 3.

`x=(15-13)/(2) \\`

• Add
  `-(13)/(2)\\` to
  `(15)/(2)\\` by finding a common denominator and adding the numerators. Then reduce the fraction to its lowest terms if possible.

`\large\underline{ \underline{ \sf \: x=1 }}`

• The system is now solved. The value of x & y will be 1 & 3 respectively.

`\huge\boxed{ \boxed{\bf \: x=1, \: y=3 }}`