Question 198300
Make an algebra problem
{{{.10a+.30b=60}}}
{{{.15a+.40b=88}}}

Solve for a variable and substitute
{{{.10a+.30b=60}}}--->{{{a=600-3b}}}
{{{.15(600-3b)+.40b=88}}}
{{{90-.45b+.40b=88}}}
{{{2=.05b}}}
{{{b=40}}}

Substitute
{{{.15a+.40(40)=88}}}
{{{.15a+16=88}}}
{{{.15a=72}}}
{{{a=480}}}

Check
{{{.10(480)+.30(40)=60}}}
{{{48+12=60}}}
{{{60}}}

grams of each of the two alloys should be used to make an alloy that contains 60g of gold and 88g of lead
480g of 10/15
40g of 30/40