Question 258874
We have the following equations
(i) {{{x=-4}}} 
(ii) {{{5x^2=4y+3x^2-8 }}}
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step 1 - substitute -4 for x and get
(iii) {{{5*(-4)^2 = 4y + 3*(-4)^2-8}}}
and then
(iv) {{{80 = 4y + 48 -8}}}
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step 2 - solve for y to get
(v) {{{80 = 4y + 40}}}
subtract 40 to get
(vi) {{{4y = 40}}}
divide to get
{{{y = 10}}}