`a+2b=3`

`a^2+b^2=26`

To solve, apply substitution method. To do so, solve for a in the first equation.

`a+2b=3`

`a+2b-2b=3-2b`

`a=3-2b`

Then, plug-in this to the second equation.

`a^2+b^2=26`

`(3-2b)^2+b^2=26`

`9-6b-6b+4b^2+b^2=26`

`5b^2-12b+9=26`

Set one side equal to zero.

`5b^2-12b+9-26=26-26`

`5b^2-12b-17=0`

Then, factor.

`(5b-17)(b+1)=0`

Set each factor equal to zero and solve for b.

`5b-17=0` and ` b+1=0`

`b=17/5` `b=-1`

Next, solve for a. To do so, plug-in the values of b to a=3-2b.

`b=-1, a=3-2(-1)=3+2=5`

`b=17/5, a=3-2(17/5)=3-34/5=15/5-34/5=-19/5` **Hence, the solutions to the system of equations are:****> `a=5` , `b=-1` ****and**

> `a=-19/5` 5, `b =17/5`

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