Answer:
[tex]\large x=3\ and\ y=-1\ or\ x=-\dfrac{23}{7}\ and\ y=\dfrac{15}{7}[/tex]
Step-by-step explanation:
[tex]\left\{\begin{array}{ccc}2x^2-y^2=17\\x+2y=1&\text{subtract 2y from both sides}\end{array}\right\\\\\left\{\begin{array}{ccc}2x^2-y^2=17&(1)\\x=1-2y&(2)\end{array}\right\\\\\text{Substitute (2) to (1):}\\\\2(1-2y)^2-y^2=17\qquad\text{use}\ (a-b)^2=a^2-2ab+b^2\\\\2(1^2-2(1)(2y)+(2y)^2)-y^2=17\\\\2(1-4y+4y^2)-y^2=17\qquad\text{use the distributive property}\\\\(2)(1)+(2)(-4y)+(2)(4y^2)-y^2=17\\\\2-8y+8y^2-y^2=17\\\\(8y^2-y^2)-8y+2=17\qquad\text{subtract 17 from both sides}\\\\7y^2-8y-15=0[/tex]
[tex]7y^2+7y-15y-15=0\\\\7y(y+1)-15(y+1)=0\\\\(y+1)(7y-15)=0\iff y+1=0\ \vee\ 7y-15=0\\\\y+1=0\qquad\text{substract 1 from both sides}\\\boxed{y=-1}\\\\7y-15=0\qquad\text{add 15 to both sides}\\7y=15\qquad\text{divide both sides by 7}\\\boxed{y=\dfrac{15}{7}}[/tex]
[tex]\text{Put the values of y to (2):}\\\\for\ y=-1\\\\x=1-2(-1)\\x=1+2\\\boxed{x=3}\\\\for\ y=\dfrac{15}{7}\\\\x=1-2\left(\dfrac{15}{7}\right)\\x=1-\dfrac{30}{7}\\x=\dfrac{7}{7}-\dfrac{30}{7}\\\boxed{x=-\dfrac{23}{7}}[/tex]
