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By using substitution: 15x+3y=5 5x+y=-6 Show me how to do this.
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Let's manipulate the second equation, we'll get it in terms of y=, and then we'll PLUG it into the first equation. \[\large 5x+y=-6\]Subtracting 5x from both sides gives us,\[\large y=\color{cadetblue}{-6-5x}\]We want to plug this in for the y in the other equation.\[\large 15x+3\color{cadetblue}{y}=5\] Which gives us,\[\huge 15x+3\color{cadetblue}{(-6-5x)}=5\]
We SUBSTITUTED one equation into the other one. From here it's not too bad. We just have to multiply the 3 by each term in the brackets, and then solve for x. If you're still confused, let me know.
Okayy I will,
@nickersia Oh thanks.
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