Question

How do I solve 2m + n = 2 and 3m - 2n = 3 using substitution?

Answer 1

`\huge \boxed{\mathbb{QUESTION} \downarrow}`

• Solve 2m + n = 2 and 3m - 2n = 3 using substitution.

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

We can use the substitution method to solve linear equations of this form. Let's solve for m & n.

`\left. \begin{array} { l } { 2 m + n = 2 } \\ { 3 m - 2 n = 3 } \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.

`2m+n=2, \: 3m-2n=3`

Choose one of the equations and solve it for m by isolating m on the left-hand side of the equal sign.

`2m+n=2`

Subtract n from both sides of the equation.

`2m=-n+2`

Divide both sides by 2.

`m=(1)/(2)\left(-n+2\right) \\`

Multiply 1/2 times -n+2.

`m=-(1)/(2)n+1 \\`

Substitute
`-(n)/(2)+1\\` for m in the other equation, 3m-2n=3.

`3\left(-(1)/(2)n+1\right)-2n=3 \\`

Multiply 3 times
`-(n)/(2)+1\\`.

`-(3)/(2)n+3-2n=3 \\`

Add
`-(3n)/(2)\\` to -2n.

`-(7)/(2)n+3=3 \\`

Subtract 3 from both sides of the equation.

`-(7)/(2)n=0 \\`

Divide both sides of the equation by
`-(7)/(2)\\`, which is the same as multiplying both sides by the reciprocal of the fraction.

`\large \underline{\underline{ \bf \: n=0 }}`

Substitute 0 for n in
`m=-(1)/(2)n+1\\`. Because the resulting equation contains only one variable, you can solve for m directly.

`\large \underline{ \underline{\bf \: m=1 }}`

The system is now solved.

`\huge \boxed{ \boxed{ \bf \: m=1, \: n=0 }}`